# Springback Analysis in Sheet Metal Forming Using Modified Ludwik Stress-Strain Relation

Authors: Sanjay Kumar Patel; Radha Krishna Lal; J. P. Dwivedi; V. P. Singh
Journal: ISRN Mechanical Engineering
 (2013)

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## Abstract

This paper deals with the springback analysis in sheet metal forming using modified Ludwik stress-strain relation. Using the deformation theory of plasticity, formulation of the problem and spring back ratio is derived using modified Ludwik stress strain relationship with Tresca and von Mises yielding criteraia. The results have been representing the effect of different value ofY/E or σo/E ratio, different values of Strain hardening index (n), Poisson’s ratio (ν), and thickness on spring back ratio (Ro/Rf). The main aim of this paper is to study the effects of the thickness, Y/E ratio, n and Poisson’s ratio in spring back ratio.


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## Body

## 1. Introduction

In sheet metal working, sheets are deformed to cylindrical and helical shapes by plastic bending with the help of a punch and die set. When such bending is properly done, the inside contour of the section matches the surface contour of the die during the forming operation. However, after the release of the applied loads, the contour assumes a different shape than that of the die because of the release of elastic stresses in the metal. This elastic distortion is commonly called springback. Springback complicates tool design in that the die must be designed to compensate for it. Consequently, it is desirable to have a method of quantitatively predicting the magnitude of springback as a function of the properties of the material and geometry involved in the forming operation.In bending, the springback is a measure of elastic recovery of radius on removal of the applied bending moment or load after the bending section is beyond elastic limit. While designing the die set, the springback factor should be taken into account to avoid mismatch while assembling formed different sections.Initially, springback studies are with sheet bending operations. Sachs [1], Schroeder [2], Gardiner [3], Singh and Johnson [4], and others studied the springback considered bending of sheets of different shapes and depicted springback as a function of material thickness, length, and width of the sheets taken. Their studies were limited to V- and U-shaped dies for applying bending loads and they predicted the springback as a measure of change in the curvature distribution. Huth [5], Nadai [6], and Upadhyay [7] have all done a number of excellent work on the elasto-plastic torsion of bars with rectangular sections, but their work has been limited to monotonically increasing loads only. Dwivedi et al. [8, 9] analytically predicted the residual angle of twist and torque relation, and so forth, for bars of elastic strain-hardening materials with narrow rectangular sections. This work, however, has the limitation that it is valid for thin rectangular strips only. Dwivedi et al. [10, 11] dealt with the torsional springback of square-section bars of linear and nonlinear work-hardening materials. Dwivedi et al. [12] also dealt with the torsional springback of L-shaped section bars of nonlinear work-hardening materials.An accurate analysis of springback has been made in the past on sheet bending and tube bending operations through experiment [13–18]. Torsional springback in thin tubes with nonlinear work hardening analysis was made by Choubey et al. [19–21].In the following, approximation equations are derived in an attempt to provide a quantitative method with practical utility for predicting the springback behavior in bent section of sheet metal as a function of die radius, sheet thickness, and stress-strain characteristics of the material.


## 2. Basic Equations and Formulation

In this paper, springback prediction approach using the modified Ludwik stress-strain relation was studied. The assumption of narrow beam of a wide sheet can significantly alter the magnitude of the predicted results in sheet metal bending because this ignorance of the transverse stresses are present during forming. In this paper, it is assumed that the bend section is a wide sheet of an elastic plastic strain hardening material in an effort to obtain more accurate expression for predicting the springback behavior. In this method, use of applied moment and curvature relation during the formation of a wide sheet of metal around a portion of cylindrical die was done to derive springback relationship. From the bending equation, we directly relate the bending moment and curvature during the forming as follows:(1)MI=σy=ER,MEI=1R,M=C*1R.From this equation, the springback is related to the applied bending moment and curvature during the formation. The modified Ludwik stress-strain relation (Figure2) is given by
(2)σ=[Eεε≤YEY(EεY)nε⩾˙YE],
where Y=σo, σ=Eε is the elastic part and σ=σo(Eε/σo)n is the plastic part, and the equation gives the total stress.


## 3. Assumption for Derivation of Springback 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 thickness of the sheet so radial stresses are assumed to be negligible.
(7)
The circumferential strains are sufficiently small so that the conventional strain and the true strain are approximately 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.


## 4. Derivation of Springback Equation for Tresca Yield Criteria

Figure1 shows the schematic plot of applied bending moment versus curvature during the formation of a wide sheet of metal around a portion of a cylindrical die at point A, the material yields and plastic deformation continuous until the inside surface of the material conforms to the inside surface of the material conforms to die at point B when the applied moment is released elastic springback occurs from B to C. The change in curvature due to this elastic springback is (1/Ro)-(1/Rf)Figure 1
Schematic representation of curvature versus applied moment during bending.Figure 2
Empirical stress-strain curve for elastic plastic material in modified ludwik stress-strain relation.From Figure3,
(3)1Ro-1Rf=Mmax∂ME/∂(1/R).Figure 3
Direction of principal stresses and curvature during bending.Figure3 shows a section of bent sheet and the coordinate system adopted. Then calculation of maximum applied bending moment by balancing the external and internal moments is as follows:
(4)Mmax=∫-t/2+t/2σxydy=2∫0t/2σxydy.Before proceeding, it is desirable to attempt to characterize the stress-strain behaviour of material in a simple tension. From the modified Ludwik stress-strain relation, we get(5)σ=[Eεε≤YEY(EεY)nε⩾˙YE].This is the equation for springback for Modified Ludwik stress-strain relation.In elastic region,(6)σ=Eε.
In plastic region, where Y=σo,
(7)σ=σo(Eσo)nεn.
Let
(8)σo(Eσo)n=K.
Then
(9)σ=Kεn.
At yield point, the stress in elastic and plastic region is equal; so we can write
(10)Eεo=Kεon,εo=(KE)1/(1-n).
From (10), stress in yield point is
(11)σo=Kεon=K(KE)n/(1-n).
It will be assumed for stresses less than σo that the stress-strain behaviour of the material in a simple tension test is elastic (6) and that yielding occurs at σo and (7) is applicable for stresses greater than σo.For combined states of stress, the relationship between principal stresses and strain for the elastic.By Hooke’s law,(12)εx=1E(σx-ν(σy+σz)),εy=1E(σy-ν(σx+σz)),εz=1E(σz-ν(σy+σx)).From assumptions that the radius of bending is large compared to thickness of the sheet so radial stresses can assume to be negligible, and the transverse strain is zero at any point in the sheet,(13)σy=εz=δz=0.And the circumferential strain for any fiber does not vary along the bent section.So,(14)εx=δx.Then, from putting the value of (12) in (13),
(15)εz=1E(σz-ν(σy+σx)),0=1E(σz-ν(σy+σx)),1E(σz-ν(σx))=0,σz=νσx.
From maximum shear stress theory of failure (Tresca yield criteria),
(16)σo2=σx-σz2,σo=σx-νσx,σo=σx(1-ν).Substituting (16) in (11),
(17)σo=σx(1-ν)=K(KE)n/(1-n),σx=K(K/E)1/(1-n)(1-ν).Then,(18)εx=1E(σx-ν(σy+σz)),εx=1E(σx-ν2σx),εx=(1E)σx(1-ν2).So in yield point, the circumferential strain is(19)εxo=(1E)σo(1-ν2).Putting yield point stress value from (11), we get
(20)εxo=K(K/E)n/(1-n)E(1-ν)(1-ν2),εxo=(KE)1/(1-n)(1+ν).
This is the approximate value of circumferential strain at elastic plastic interface. Then in elastic region, the circumferential stress is
(21)σx=E(1-ν2)εx.
Here,   εx=y/Ro.Then,(22)σx=E(1-ν2)(yRo),
for 0≤εx≤(K/E)1/(1-n)(1+ν).From the deformation theory of plasticity,(23)δx=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σx-σy2-σz2),δy=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σy-σx2-σz2),δz=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σz-σx2-σy2).From assumptions that the radius of bending is large compared to thickness of the sheet so radial stresses can assume to be negligible and the transverse strain is zero at any point in the sheet,σy=εz=δz=0.And the circumferential strain for any fiber does not vary along the bent section. So,εx=δx.Put this value in (23). We get
(24)δz=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σz-σx2-σy2),0=1K1/n(σx2+σz2-σzσx)(1-n)/2n(σz-σx2).So,(25)0=(σz-σx2),σz=σx2.Putting the value ofσz into (23), we found that
(26)δx=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σx-σy2-σz2),δx=1K1/n(σx2+σx24-σx22)(1-n)/2n(σx-σx4),δx=1K1/n(3σx24)(1-n)/2n(3σx4),δx=1K1/n(34)(1+n)/2nσx1/n,σx=K(3/4)(1+n)/2δxn.From initial assumption that the circumferential strain are sufficiently small so that the conventional strain and true strain are approximately equivalent,(27)εx=yRo=δx.Putting the value ofδx into (26), we get
(28)σx=K(3/4)(1+n)/2(yRo)n,
where σx is valid for plastic region; that is,
(29)(KE)1/(1-n)(1+ν)≤εx≤t2.From (4), we found maximum bending moment is
(30)Mmax=2[∫0t/2σxydy],
where we can split the limit into elastic or plastic part, from 0 to Roex elastic region and Roex to (t/2) plastic region, where Roex is calculated by
(31)Roex=Ro(KE)1/(1-n)(1+ν).
Now,
(32)Mmax=2[∫0Roexσx(elastic)ydy+∫Roext/2σx(plastic)ydy]=2[∫0Roex(E1-ν2)y2dyhhhhhh+∫Roext/2K(3/4)(1+n)/2(yRo)nydy]=2[∫0Roex(E1-ν2)y2dyhhhhhh+∫Roext/2K(3/4)(1+n)/2(1Ro)nyn+1dy]=2(1Ro)[(E1-ν2)y33]0Roexhhhhhh+2[K(3/4)(1+n)/2(1Ro)nyn+2n+2]Roext/2.
Put the value of Roex in above equation.Then(33)=2[ERo2(K/E)3/(1-n)3(1-ν)(1+ν)2]+2[K(t/2)n+2(3/4)(1+n)/2Ron(n+2)hhhhhhh-KRo2(K/E)(n+2)/(1-n)(1+ν)n+2(3/4)(1+n)/2(n+2)].From elementary plate theory [17] applied moment per unit width during elastic unloading,
(34)MEI=σxy,ME=2·E·(t/2)33(1-ν2)·R.The slope of elastic recovery in elastic unloading is(35)∂ME∂(1/R)=2·E·(t/2)33(1-ν2).Putting this into (3), we get
(36)1Ro-1Rf=Mmax∂ME/∂(1/R)RoRf=1-Ro(Mmax∂ME/∂(1/R)),RoRf=1-Ro{(2Rot)3(KE)3/(1-n)(1+ν)3hhhhhhh+3K(1-ν2)(2Ro/t)1-nE(3/4)(1+n)/2(n+2)hhhhhhh-3((2Ro)/t)3(K/E)3/(1-n)(1+ν)n+3(1-ν)(3/4)(1+n)/2(n+2)},RoRf=1-3K(1-ν2)E(3/4)(1+n)/2(n+2)(2Rot)1-n+[(2Rot)(KE)1/(1-n)(1+ν)]3×[3(1+ν)n(1-ν)(3/4)(1+n)/2(n+2)-1],RoRf=1-3(1-ν2)(3/4)(1+n)/2(n+2)(Eσo)n-1(2Rot)1-n+[(2Rot)(σoE)1(1+ν)]3×[3(1+ν)n(1-ν)(3/4)(1+n)/2(n+2)-1].This is the equation for springback ratio for modified Ludwik stress-strain relationship using maximum shear stress theory.


## 5. Derivation of Springback Equation for Von Mises Yielding

From maximum shear stress theory of failure (von Mises yielding criteria),(37)σo2=12{(σx-σy)2+(σy-σz)2+(σz-σx)2},σo2=12{σx2+σz2+σz2-2σzσx+σx2},σo=σx(1+ν2-ν)1/2,
by using (13) and (14).Substituting (15) in (11), at yield point,
(38)σo=K(KE)n/(1-n)=σox(1+ν2-ν)1/2,σox(1+ν2-ν)1/2=K(KE)n/(1-n),σxo=K(K/E)n/(1-n)(1-ν+ν2)1/2.Now,(39)εx=1E(σx-ν(σy+σz)),εx=1E(σx-ν2σx),εx=1Eσx(1-ν2).So, in yield point, the circumferential strain is(40)εxo=1Eσox(1-ν2).Putting yield point stress value from (38) in (40), we get
(41)εxo=K(K/E)n/(1-n)E(1-ϑ+ϑ2)1/2(1-ν2),εxo=(KE)1/(1-n)(1-ν2)(1-ν+ν2)1/2.This is the approximate value of circumferential strain at elastic plastic interface.Then in elastic region, the circumferential stress is(42)σx=E(1-ν2)εx.
Here, εx=y/Ro.Putting value of circumferential strain, we get(43)σx=E(1-ν2)(yRo),for0≤εx≤(K/E)1/(1-n)((1-ν2)/(1-ν+ν2)1/2).From the deformation theory of plasticity [16],
(44)δx=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σx-σy2-σz2),δy=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σy-σx2-σz2),δz=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σz-σx2-σy2).From assumptions that the radius of bending is large compared to thickness of the sheet so radial stresses can assume to be negligible and the transverse strain is zero at any point in the sheet,(45)σy=εz=δz=0.
And the circumferential strain for any fiber does not vary along the bent section.So,(46)εx=δx.
Put these values in (44). We get
(47)δz=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σz-σx2-σy2),0=1K1/n(σx2+σz2-σzσx)(1-n)/2n(σz-σx2).So,(48)0=(σz-σx2),σz=σx2.
Putting the value of σz into (44), we get
(49)δx=1K1/n(σx2+σy2+σz2-σxσy-σyσz-σzσx)(1-n)/2n×(σx-σy2-σz2),δx=1K1/n(σx2+σx24-σx22)(1-n)/2n(σx-σx4),δx=1K1/n(3σx24)(1-n)/2n(3σx4),δx=1K1/λ(34)(1+n)/2nσx1/n,σx=K(3/4)(1+n)/nδxn.From initial assumption that the circumferential strain are sufficiently small so that the conventional strain and true strain are approximately equivalent,(50)εx=yRo=δx.
Putting the value of δx into (49), we get
(51)σx=K(3/4)(1+n)/n(yRo)n,
where σx is valid for plastic region; that is,
(52)for(KE)1/(1-n)(1-ν2)(1-ν+ν2)≤εx≤t2.
From (4), maximum bending moment is
(53)Mmax=2[∫0t/2σxydy].
We can split the limit into elastic or plastic part from 0 to t/2 to 0 to Roex elastic region and Roex  to t/2 plastic region where Roex  is calculated by
(54)Roex=(KE)1/(1-n)Ro(1-ν2)(1-ν+ν2)1/2,Mmax=2[∫0Roexσx(elastic)ydy+∫Roext/2σx(plastic)ydy]=2[∫0RoexE1-ν2y2(1Ro)dyhhhhh+∫Roext/2K(3/4)(1+n)/2(yRo)nydy]=2[∫0Roex(E1-ν2)y2(1Ro)dyhhhhh+∫Roext/2K(3/4)(1+n)/2(1Ro)nyn+1dy]=2(1Ro)[(E1-ν2)y33]0Roex+2[K(3/4)(1+n)/2(1Ro)nyn+2n+2]Roext/2.
Putting the value of Roex, we get
(55)Roex=(KE)1/(1-n)Ro(1-ν2)(1-ν+ν2)1/2.
Then
(56)Mmax=2[ERo3(K/E)3/(1-n)((1-ν2)3/(1-ν+ν2)3/2)3(1-ν2)Ro]+2[K(t/2)n+2(3/4)(1+n)/2Ron(n+2)hhhh-KRo(n+2)(K/E)(n+2)/(1-n)(1-ν2)n+2(3/4)(1+n)/2Ron(n+2)(1-ν+ν2)(n+2)/2].From elementary plate theory [17] applied moment per unit width during elastic unloading,
(57)ME=2·E·(t/2)33(1-v2)·R.
Now, the slope of elastic recovery in elastic unloading is
(58)∂ME∂(1/R)=2·E·(t/2)33(1-ν2).
Putting this into (3), we get
(59)1R0-1Rf=Mmax∂ME/∂(1/R),RoRf=1-RoMmax∂ME/∂(1/R),RoRf=1-3K(1-ν2)E(n+2)(3/4)(n+1)/2(2Rot)1-n+[(2Rot)(KE)1/(1-n)]3×[(1-ν2)3(1-ν+ν2)3/23(1-ν2)n+3(3/4)(n+1)/2(n+2)(1-ν+ν2)(n+2)/2hhHhh-(1-ν2)3(1-ν+ν2)3/2],RoRf=1-3(1-ν2)(n+2)(3/4)(n+1)/2(σoE)1-n(2Rot)1-n+[(2Rot)(σoE)]3×[3(1-ν2)n+3(3/4)(n+1)/2(n+2)(1-ν+ν2)(n+2)/2hhhhhy-(1-ν2)3(1-ν+ν2)3/2].
This is the equation for springback ratio for modified Ludwik stress-strain relationship using von Mises yield criteria.


## 6. Results and Discussion

For the sheet metal bending, the springback is calculated from derived equation (using modified Ludwik stress-strain relationship with Tresca and von Mises yield criteria).(60)RoRf=1-3(1-ν2)(3/4)(1+n)/2(n+2)(Eσo)n-1(2Rot)1-n+[(2Rot)(σoE)1(1+ν)]3×[3(1+ν)n(1-ν)(3/4)(1+n)/2(n+2)-1],RoRf=1-3(1-ν2)(n+2)(3/4)(n+1)/2(σoE)1-n(2Rot)1-n+[(2Rot)(σoE)]3×[3(1-ν2)n+3(3/4)(n+1)/2(n+2)(1-ν+ν2)(n+2)/2hhhhh-(1-ν2)3(1-ν+ν2)3/2].
Here, from (8),
(61)K=[E(Eσo)n-1].Then,K/E=(σo/E)1-n.We can see that both equations have a last term that is very negligible because theσo/E  or  Y/E=5.5×10-4 and its three times will be 10-12; therefore, we have to neglect the last term of both equations.Then we write,(62)RoRf=1-3(1-ν2)(3/4)(1+n)/2(n+2)(EY)n-1(2Rot)1-n,(63)RoRf=1-3(1-ν2)(n+2)(3/4)(n+1)/2(YE)1-n(2Rot)1-n.
Here, we can see that both (62) and (63) are the same if we can apply Tresca or von Mises, getting same result but last term is different. The values of last term are much less; therefore, we neglect this term.Therefore, we can say that if we apply Tresca or von Mises yield criteria, then we are getting same result. This equation is dependent on theRo/t, σo/E, strain hardening coefficient n, and ν is Possion’s ratio.Figure4 (n=0.1), Figure 5 (n=0.2), Figure 6 (n=0.3), and Figure 7 (n=0.4) show the variation in springback ratio along with change in Ro/t with different σo/E ratio. The higher value of n (strain hardening coefficient) material to closer to elasto-ideal plastic material, and thickness is one mm, and σo/E value is 5.5×10-4,1.522×10-3, and 2.4×10-3 (for different material c10100 copper, 1100al, and 1065steel). From the figure, we come to know that as the with increasing ratio of Ro/t is decrease the springback ratio and with decreasing σo/E springback ratio is increases.Figure 4
Springback ratio withRo/t with different values of σo/E or Y/E at n=0.1, ν=0.33.Figure 5
Springback ratio withRo/t with different values of σo/E or Y/E at n=0.2, ν=0.33.Figure 6
Springback ratio withRo/t with different values of σo/E or Y/E at n=0.2, ν=0.33.Figure 7
Springback ratio withRo/t with different values of σo/E or Y/E at n=0.4, ν=0.33.Figures8, 9, and 10 show the variation in springback ratio along with Ro/t with the different value of n with fixed value of σo/E=5.5×10-4, σo/E=1.522×10-3, and σo/E=2.4×10-3, here we take the values of n=0.1,0.2,0.3, and 0.4. From the figure, we come to know that with increasing value of n, springback ratio is decreasing.Figure 8
Springback ratio withRo/t with different values of n at Y/E=5.5×10-4, ν=0.33.Figure 9
Springback ratio withRo/t with different values of n at Y/E=1.5×10-3, ν=0.33.Figure 10
Springback ratio withRo/t with different  values of n at Y/E=2.4×10-3, ν=0.33.Figures11, 12, and 13 show the variation in springback along with Ro/t with different value of Possion’s ratio ν=0.25, ν=0.35, and ν=0.45 with fixed  σo/E=5.5×10-4, σo/E=1.522×10-3, σo/E=2.4×10-3, and n=0.3. From the figure, we come to know that with increasing value of ν, springback ratio is increasing.Figure 11
Springback ratio withRo/t with different values of ν at Y/E=5.5×10-4.Figure 12
Springback ratio withRo/t with  different values of ν at Y/E=1.5×10-3.Figure 13
Springback ratio withRo/t with different values of ν at Y/E=2.4×10-3.Figures14, 15, and 16 show the variation in springback along with t with different value of n=0.1, 0.2, 0.3, and 0.4 and where σo/E=5.5×10-4, σo/E=1.522×10-3, σo/E=2.4×10-3, ν=0.33, and Ro=40 is fixed. From the figure, we come to know that the springback ratio rapidly increases in range of 0 to 2 mm thickness, then the increase in springback ratio is stable.Figure 14
Springback ratio witht for different values of n at Y/E=5.5×10-4.Figure 15
Springback ratio witht for different values of n at Y/E=1.5×10-3.Figure 16
Springback ratio witht for different values of n at Y/E=2.4×10-3.


## 7. Conclusions

Based on the results presented in chapter of results and discussion, following conclusions about the springback analysis in sheet metal forming using nonlinear constitutive equation, the following applies.(i)
The theoretical analysis for the sheet metal under pure bending has been done, and it was found that the prediction of springback is quite successfull.(ii)
As expected, elastic recovery is found to be more with decreasing values of work hardening coefficient. At lower values of n, the material will approach to an elasto-ideally plastic behavior.(iii)
Springback ratio increases with increasing thickness.(iv)
Springback ratio increases with decreasing ratio of yield point stress to Young’s modulus of elasticity(v)
Springback ratio is increasing with increasing  Possion’s ratio.(vi)
Springback ratio for Tresca and von Mises yield criteria, we say thatRo/t is less than 20, there will be slightly change in that is negligible and for high value of Ro/t or more than 20, differences in springback ratio is increases.



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*Source: 640958-2013-11-03.xml*