Numerical and experimental analysis of elastic–plastic pure bendingand springback of beams of asymmetric cross-sections

M. Sitar, F. Kosel, M. Brojan n

Laboratory for Nonlinear Mechanics, Faculty of Mechanical Engineering, University of Ljubljana, Askerceva 6, SI-1000 Ljubljana, Slovenia

a r t i c l e i n f o

Article history:Received 24 July 2014Received in revised form21 October 2014Accepted 1 November 2014Available online 7 November 2014

Keywords:Elastic–plastic bendingAsymmetric cross-sectionSpringbackElastic modulus evolutionExperiments

a b s t r a c t

We present a procedure for numerical computation of elastic–plastic bending and springback of beamswith asymmetric cross-sections. Elastic-nonlinear hardening behavior of the material is assumed andboth isotropic and kinematic hardening models are considered. The strains are described as a function ofrotation and shift of the neutral axis and the curvature of the beam. Exact geometric expressions forlarge deflections and large rotations are taken into account during bending process. A complete loadinghistory is taken into account including the effect of the local loading during the monotonic decrease ofthe load. Numerical examples confirm a strong influence of the load on the final and springback rotationof the neutral axis, its shift, and curvature of the beam for different cross-sections and materials.A custom made forming tool was designed and manufactured in-house to experimentally evaluate theproposed solution procedure. It is shown that relative difference between experimentally andtheoretically predicted results of the final radius of curvature of the formed beam is 0.17770.683%,if also the effect of pre-strain on elastic modulus is taken into consideration.

& 2014 Published by Elsevier Ltd.

1. Introduction

Either as vital parts of load-bearing structures in mechanicaland civil engineering or merely as an aesthetic feature in archi-tecture, curved beams are most commonly made via some sort offorming process. V-bending, roll-bending, air and edge-bending,hydroforming, etc. are some of the examples of technological/manufacturing processes for obtaining the desired shape. Theprediction of the (final) shape can be a complex task, especiallybecause real-life materials often exhibit nonlinear mechanicalresponse to loading.

In the forming process, the material undergoes elastic–plasticdeformations. The plastic part of deformation changes the originalshape of the object permanently, whereas the elastic part returnsthe deformed shape back towards initial configuration. Since acertain amount of elastic deformation is practically always present,the final shape of the object is not the same as the shape of theforming tool itself. A common way to deal with this problem is toadd special techniques to reduce the effect of elastic recovery (alsoknown as springback), such as extra features in radii, using smallerradii, or varying blankholder force in the forming process. Thesetechniques reduce the effect of springback, but the formed partwill always tend to springback by a certain amount.

In the available literature one can find a considerable numberof papers devoted to this subject. Kosel et al. [1] presented ananalytical solution of the simplified model for predicting thespringback of beams made from material with an elastic-linearhardening response. The beams were subjected to repeated purebending and unbending process and complete strain history wasconsidered. The influence of axial force on the bending andspringback of the elastic–ideal plastic beam was investigated byYu and Johnson [2]. Johnson and Yu [3] developed formulas forspringback of beams and plates undergoing linear work hardening.Springback of equal leg L-beams subjected to elastic–plastic purebending was described by Xu et al. [4]. A theoretical model topredict the final geometrical configurations of wires made ofdifferent materials after loading and unloading was proposed byBaragetti [5]. Although analytical solutions can be obtained onlyfor relatively simple problems. Their advantage is that they enablebetter insight and understanding of the problem and the influenceof the process parameters. For more complex problems, however,the general practice is to refer to numerical techniques. Thus Liet al. [6] analyzed draw-bend tests of sheet metals using finiteelement modeling, where some of the results have been comparedwith experiments. The error associated with numerical through-thickness integration was investigated by Wagoner and Li [7]. Theprediction model for springback in a wipe-bending process wasdeveloped by Kazan et al. [8] using artificial neural networkapproach together with the finite element method. Panthi et al.[9] analyzed and examined the effect of load on springback of a

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International Journal of Mechanical Sciences

http://dx.doi.org/10.1016/j.ijmecsci.2014.11.0060020-7403/& 2014 Published by Elsevier Ltd.

n Corresponding author. Tel.: þ386 1 4771 604; fax: þ386 1 2518 567.E-mail address: miha.brojan@fs.uni-lj.si (M. Brojan).

International Journal of Mechanical Sciences 90 (2015) 77–88

typical sheet metal bending process using a large deformationalgorithm. Furthermore, Ragai et al. [10] investigated the influenceof sheet anisotropy on the springback of drawn-bend specimensby means of experiments and finite element analysis. Vladimirovet al. [11] developed a finite strain model by combining bothnonlinear isotropic and kinematic hardening, where for theintegration of the equations a new algorithm based on anexponential map was used. An interesting phenomenon, adecrease of elastic modulus, can be observed in experiments on(e.g. metal) materials during plastic deformation. Ghaei [12]presented a numerical procedure which took into account alsothis effect. He implemented the elasto-plastic constitutive lawsassuming elastic modulus as a function of effective plastic strain.The evolution of the elastic modulus with plastic deformation wasalso studied in other papers, see e.g. [13–20]. The change of elasticmodulus during unloading is usually described by introducing thelinear chord modulus [14,19,20] in many practical applications.However, experimental studies have shown that the elastic defor-mation during unloading is not perfectly linear. In work done bySun and Wagoner [18] a new concept of quasi-plastic-elastic (QPE)strain was introduced within the continuum framework to modelthe nonlinear unloading behavior. It was shown that QPE conceptis superior to the linear chord modulus in accurate prediction ofspringback in cases when the unloading stops at non-zero stresses.

These studies of elastic–plastic deformations of beams aremostly limited to symmetric cross-sections. Here, we present thesolution procedure for elastic–plastic bending and springback ofbeams with asymmetric cross-sections. Elastic-nonlinear harden-ing behavior of the material is assumed and both isotropic andkinematic hardening models are considered. The strains aredescribed as a function of rotation and shift of the neutral axisand the curvature of the beam. Exact geometric expressions forlarge deflections and large rotations are used during bendingprocess. Strains, on the other hand, are considered to remainsmall. A complete loading history, including the effect of the localloading during the monotonic decrease of the load, is taken intoaccount. Numerical examples confirm a strong influence of theload on the final and springback rotation of the neutral axis, itsshift, and curvature of the beam for different cross-sections andmaterials. Generally, forming of beams with asymmetric cross-sections involves also torsional deformations, which we do notconsider in our computations. Instead we find a special combina-tion of forming parameters to constrain (to remove the effect oftorsion from) an asymmetric rectangular L-beam to deform in oneplane. Note that the presented solution procedure can easily beused for more complex shapes of cross-sections. We also present acustom-made forming tool, designed and manufactured in-houseto experimentally evaluate the proposed solution procedure.Practically perfect planar shapes of the beams are obtained afterforming, showing an excellent agreement with theoretical predic-tions, especially when the effect of the pre-strain on elasticmodulus is considered.

2. Formulation of the problem

We consider a beam of asymmetric cross-section subjectedto the bending moment M(t) in direction α, as shown in Fig. 1,and assume an isotropic, homogeneous material which exhibitselastic-nonlinear hardening behavior. The yield point is definedby non-negative parameters σ0 and ε0 (cf. Fig. 3). Suppose thatthe beam is stress-free before loading and that the mechanicalresponse can be described by

σðε; ε0;σ0Þ ¼f eðεÞ for jεjrε0

f pðε; ε0;σ0Þ for jεj4ε0;

(ð1Þ

where f eðεÞ and f pðε; ε0;σ0Þ represent stress–strain response in theelastic and plastic regions, respectively. Both isotropic and kine-matic hardening models are considered.

Exact geometric expressions for large deflections and largerotations are taken into account during bending process. Followingthe Euler–Bernoulli theory, valid for slender beams, a straindistribution over the cross-sectional area due to bending can bedescribed by the following expression:

ε¼ �κz¼ �κðzC�zsÞ¼ �κð�ðy�YCÞ sinβþðz�ZCÞ cosβ�zsÞ; ð2Þ

where κ ¼ 1=r is the curvature of the beam (and r is its radius ofcurvature), β is the rotation of the neutral axis, zs is the shift of theneutral axis from the centroid and YC, ZC are the coordinates of thecenterline taken from the reference coordinate system y–z (seeFig. 1). The neutral axis is found from the no-strain condition,ε� 0. In the case of a linear elastic beam the neutral axis goesthrough the centroid of the cross-section.

Equations of static equilibrium of the beam ∑i F!

i ¼ 0 and

∑iM!

i ¼ 0 and static equilibrium of the infinitesimal element yielddMy=ds¼ 0 and dMz=ds¼ 0 (My, Mz are constants) in the case ofpure bending, where s is a curvilinear coordinate along the lengthof the deformed beam (note that curvature κ ¼ dϑðsÞ=ds, whereϑðsÞ is the angle of inclination of the plane, tangent to the beam'sneutral surface, at the local coordinate s). The stress resultants are

N¼ZAσ dA; My ¼ �

ZAσz dA; Mz ¼

ZAσy dA; ð3Þ

where N represents inner axial force, My and Mz are inner bendingmoments in directions of y-axis and z-axis, respectively, and σ isthe normal stress. Due to the nonlinearity of the problem, thecomputation of integrals in Eq. (3) is numerical. The integrationdomain (the cross-section) is divided into n rectangular elements.A generic element Ai, iAf1;2;…;ng (cf. Fig. 1) is defined by fournodes in which the mechanical properties are known. Since acomplete loading history has to be considered and the effect of thelocal loading during the monotonic decrease of the bendingmoment occurs (and vice versa in monotonic increase of the loadin non-virgin material), the bending moment MðtÞ : 0-Mmax andMmax-0 is applied incrementally, as shown in Fig. 2. Here variablet represents a pseudo-time, which is used to follow successiveincrements of the load (and stress and strain).

The mechanical state corresponding to the current load Mðt0Þthus includes a complete loading history. Since the strain in eachnode of the cross-section εt0 �Δt

j , jAf1;2;…;nNg (where nN is the

yi,3

AC

y

z

zs

r

y yCz z C

yC

zC

YCZC

M t( )

zz

neutralaxis

Ai

i,2

i,4i

i,1

i,3i,3

i,3

zi,3

Fig. 1. Bending stress and strain state in the cross-section of the beam.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8878

total number of nodes) is known from the previous load,Mðt0�ΔtÞ, the change of strain in the current increment can becalculated from

Δεj ¼ εj�εt0 �Δtj ; ð4Þ

where strain εj (cf. Eq. (2)),

εj ¼ �κð�ðyj�YCÞ sinβþðzj�ZCÞ cosβ�zsÞ; ð5Þ

is a function of three variables, κ, β and zs. Note that we write εjinstead of εt0j for the strain in the current increment.

Using an algorithm for updating the stress–strain relations,presented in Section 2.1 and Appendix B, the change of the stresscan be calculated for each node of the cross-section:

Δσj ¼ σupj ðΔεjÞ; ð6Þ

jAf1;2;…;nNg. Unknown κ, β and zs (Eq. (5)) are computednumerically from

ΔN¼ZAΔσ dA-0¼ 1

4∑n

i ¼ 1∑4

k ¼ 1Δσi;kAi;

ΔMy ¼ �ZAΔσz dA-ΔM0 cosα¼ 1

4∑n

i ¼ 1∑4

k ¼ 1Δσi;kzi;kAi;

ΔMz ¼ZAΔσy dA-ΔM0 sinα¼ �1

4∑n

i ¼ 1∑4

k ¼ 1Δσi;kyi;kAi: ð7Þ

Quasi-Newton iterative method and trapezoidal integration ruleare used [22].

Finally, the current stress can be calculated as

σj ¼ σt0 �Δtj þΔσj; ð8Þ

where σt0 �Δtj represents the stress known from the previous

increment.

2.1. Algorithm for updating stress–strain relations

In this subsection, a computational procedure is presented forupdating the stress–strain relations of the j-th fiber. The isotropichardening model is considered. Based on diagrams in Fig. 3,updated σup

j ðεupj Þ, which are used to find the change of the stress

Δσj between the current and previous load, can be described bythe modified stress–strain relation of the virgin materialσmodj ðεmod

j Þ shifted by the stress σsj and strain εsj according to the

σupj –εupj coordinate system. This can be written as

σupj ðεupj Þ ¼ σmod

j ðεupj �εsj Þþσsj ; ð9Þ

σmodj ðεmod

j Þ ¼ σjðεmodj ; ε0; upj ;σ0; up

j ; εfpj ;σfpj Þ; ð10Þ

σjðεmodj ; ε0; upj ;σ0; up

j ; εfpj ;σfpj Þ

¼f eðεmod

j Þ for jεmodj jrε0; upj

f pðεmodj ; ε0; upj ;σ0; up

j ; εfpj ;σfpj Þ for jεmod

j j4ε0; upj ;

8<: ð11Þ

where an updated yield point is defined by σ0; upj and ε0; upj . With

the two additional (non-negative) parameters σfpj and εfpj (due to

the change of the plastic stress–strain response in the completeloading history), Eq. (11) represents the modified stress–strainrelation of the virgin material (defined by Eq. (1)).

Linear function f e, from Eq. (11), describes the stress–strainrelation in elastic region (jεmod

j jrε0; upj ),

f eðεmodj Þ ¼ signðεmod

j Þjεmodj jE; ð12Þ

where E represents the modulus of elasticity. The plastic responseis described by f p, valid in jεmod

j j4ε0; upj . For the purposes of ourstudy we choose the modified Ludwick law with three materialparameters EL, kL and εL (see [21,23,24] for details):

f pðεmodj ; ε0; upj ;σ0; up

j ; εfpj ;σfpj Þ ¼ sign εmod

j

� �

� σ0; upj �σfp

j þEL jεmodj jþεL� ε0; upj �εfpj

� �h i1=kL �ε1=kLL

� �� �:

ð13Þ

Here, we describe the procedure for updating the stress–strainrelations of the j-th fiber in each increment of the load. Isotropichardening is considered. Graphical description is given in Fig. 3 forthe first two increments of the load, where the total strain is thesum of the elastic and plastic part of the strain:

εj ¼ εej þεpj : ð14Þ

Pseudo-code:.

1. Start with stored known variables for the previous load, i.e. attime t0�Δt: σj, εj, σ0; up

j , ε0; upj , σfpj , ε

fpj , σ

sj , ε

sj and εpj

2. An increment of the load causes change of the strain: Δεj3. Calculate change of the stress state: Δσj ¼ σup

j ðΔεjÞ4. Calculate current stress and strain state: σj ¼ σjþΔσj;

εj ¼ εjþΔεj5. Set: σ0

j ¼ σ0; upj ; ε0j ¼ ε0; upj

6. Check for the yield condition:� If jσjj�σ0

j o0;Then Goto step 7� Else Goto step 8

7. Elastic stress state� Calculate shift of the modified stress–strain relation:σsj ¼ �σj; εsj ¼ �ðΔεj�εsj Þ

� Exit8. Elastic–plastic stress state

� Update yield point: σ0; upj ¼ jσjj; ε0; upj ¼ σ0; up

j =E� Calculate change of the plastic strain: Δεpj ¼ jΔεjj�jΔσj=Ej� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞΔεpj� Calculate change of the plastic response for a given incre-

ment of the load: Δσfpj ¼ σ0; up

j �σ0j ; Δε

fpj ¼Δεpj þΔσfp

j =E� Calculate change of the plastic response over the complete

loading history: σfpj ¼ σfp

j þΔσfpj ; ε

fpj ¼ εfpj þΔεfpj

� Calculate shift of the modified stress–strain relation:σsj ¼ �signðσjÞσ0; up

j ; εsj ¼ �signðσjÞε0; upj� Exit

3. Numerical examples

Based on the above procedure we present the results ofnumerical analysis of three examples, equal leg L-beam, T-beamand rectangular L-beam. Equilibrium (7) is satisfied within atolerance 10�10.

M t( )

t

Mmax

t

0

0 t0t0 t

M t( )

M t t( )

0

0

unloadingloading

M

Fig. 2. Loading regime M(t) and an increment of the load ΔM0.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–88 79

3.1. Equal leg L-beam

The elastic–perfectly plastic equal leg L-beam is subjected tothe bending moment M(t), which acts in the direction α¼ 1801 asshown in Fig. 4. Plastic stress–strain response is described byf pðε; ε0;σ0Þ ¼ signðεÞσ0. For the relative thickness b=a¼ 0:1 thesame problem can be found in Ref. [4]. Dimensionless parameterm¼MðtÞ=Me is introduced, where Me represents the maximumelastic bending moment. Fig. 5 shows rotation of the neutral axisβ, its shift zs and curvature of the beam κ with respect to theparameter m for different maximum bending moments describedby mmax ¼Mmax=Me, see Fig. 2. The complete loading–unloadingprocess is considered and the relative thickness is fixed atb=a¼ 0:1. Relatively large differences in rotations β and shifts zsof the neutral axis during the unloading process (when the cross-section of the beam has just been plasticized) can be observed.

The effect of mmax on the final rotation βfin and shift zfins of theneutral axis and the final curvature κfin of the beam for differentvalues of the relative thickness b=a is presented in Fig. 6. The same

effect on springback rotations βs ¼ βfin�βðmmaxÞ, shiftszss ¼ zfins �zsðmmaxÞ and curvatures κs ¼ κfin�κðmmaxÞ is depictedin Fig. 7.

In the case of mmaxr1 (elastic region) the final shifts of theneutral axis and the final curvatures of the beam remain constant,i.e. zfins =a¼ 0:0 and κfin=κe ¼ 0:0, respectively, whereas the finalrotations of the neutral axis also remain constant, but differdue to the relative thickness b=a, i.e. βfin6�30:6281, �30:1751,

�29:4491 and �27:3971 for b=a¼ 0:1, 0.15, 0.2 and 0.3, respec-tively. Here, κe represents curvature of the beam subjected to themaximum elastic bending moment Me.

In comparison to the results reported in [4], our calculationsyield slightly different results (for the final rotation compareFig. 6 from [4] with Fig. 6 here). We attribute the differences tothe solution procedure used by [4], which included a simplifiedexpression for rotations during the unloading process. For the caseof the relative thickness b=a¼ 0:2 and maximum bending momentmmax ¼ 1:9 the stress states for the fully loaded and unloaded caseare plotted in Fig. 8. The cross-sectional model included n¼7600,11,100, 14,400 and 20,400 square elements for b=a¼ 0:1, 0.15,0.2 and 0.3, respectively.

Since the stress–strain relations during the unloading differsfrom that in the loading process, the contours (representing thestress distribution over the cross-section) are no longer parallel tothe neutral axis after unloading, see Fig. 8. It can be mentionedthat rotating and shifting of the neutral axis during the unloadingprocess causes the local loading of some fibers, especially those,which are close to the neutral axis.

3.2. T-beam

A T-beam is subjected to the bending moment M(t), which actsin the direction α¼ �451, as shown in Fig. 9. The material whichconstitutes the beam is obeying the elastic-linear hardeningrheological model, as graphically described in Fig. 9. An isotropictype of strain-hardening model is considered.

The stress–strain relations of the given material can be math-ematically described by the following expression:

σðε; ε0;σ0Þ ¼ signðεÞjεjE for jεjrε0

signðεÞðσ0þEtðjεj�ε0ÞÞ for jεj4ε0;

(ð15Þ

where Et denotes tangent modulus in the plastic region. The sameas in previous numerical example, dimensionless parameterm¼MðtÞ=Me and curvature of the beam κe at maximum value ofthe elastic bending moment Me are employed. The relativethickness is fixed at b=a¼ 2=15.

Fig. 10 shows the relationships between the maximum bendingmoment and the final rotations βfin, shifts zfins and curvatures κfin

for different values of the material parameter μ¼ E=Et. Fordifferent values of μ the influence of mmax on springback rotationsβs, shifts zss and curvatures κs are presented in Fig. 11. In Fig. 12 the

0

0

yield limit

p

up. yield limit

mod

mod

updated yield limit

j

j0j

j

fpj

j

0j

0, upj

0, upj

j

fpj

0j

0, upj

j

j

j

j

0, upj

0, upj

0, upj

0, upj

sj

sj

upj

upj

j

jmod

modj

j

0, upj

0, upj

0, upj

0, upj

sj

sj

upj

upjup. yield limit

1 32

pf ( ; , )0j

0j

ef ( )j

j

modjpf ( ; , , , )0, up

j0, upj

fpj

fpj

ef ( )modj

ef ( )modj ef ( )mod

j

modjpf ( ; , , , )0, up

j0, upj

fpj

fpj

modjpf ( ; , , , )0, up

j0, upj

fpj

fpj

Fig. 3. Evolution of the updated stress–strain relation σupj ðεupj Þ of the j-th fiber, jAf1;2;…;nNg. Isotropic hardening is considered.

z

y

C M t( )yC

zC

a

a

bn.a.

Fig. 4. Equal leg L-beam.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8880

1.2 1.4 1.6 1.80

2

4

6

8

1.0 1.2 1.4 1.6 1.8

0.11

0.10

0.09

0.08

0.07

0.06

1.0 1.2 1.4 1.6 1.8

23

22

21

20

19

18

17

b a 0.1

0.15

0.2

0.3

b a 0.1

0.15

0.2

0.3

b a 0.1

0.15

0.2

0.3

(°)

fin fina

z s

efin

mmax mmax mmax

Fig. 6. Final rotations βfin, shifts zfins and curvatures κfin for different values of the relative thickness b=a.

1.0 1.2 1.4 1.6 1.8

2.0

1.8

1.6

1.4

1.2

1.0

1.0 1.2 1.4 1.6 1.8

0.10

0.08

0.06

0.04

0.02

1.0 1.2 1.4 1.6 1.80

2

4

6

8

10

12

(°)

s

mmaxmmax

sa

z s

es

0.3

0.2

0.15

b a 0.1

0.3

0.2

b a 0.10.15

1.8 1.91.9

1.8

1.85

1.85

0.20.3

0.15b a 0.1

mmax

Fig. 7. Springback rotations βs, shifts zss and curvatures κs for different values of the relative thickness b=a.

Bending stress1.00000

0.42857

0.714290.85714

0.57143

0.00000

0.285710.14286

0.285710.428570.571430.714290.857141.00000

0 0.2 0.4 0.6 0.8 1.00

0.2

0.4

0.6

0.8

1.0

az

0

y a

m mmax

0.90000

0.41555

0.657770.77889

0.53666

0.05221

0.294440.17332

0.068900.190020.311130.432240.553350.674470.79558

00

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

y a

Bending stress 0

az

m 0.0

Fig. 8. Fully loaded (left) and unloaded (right) stress states for the relative thickness b=a¼ 0:2 and mmax ¼ 1:9.

mmm

1.2 1.31.4

1.51.6

1.7

1.8

m 1.9

1.2 1.31.4

1.51.6

1.71.8max

m 1.9max

0.0 0.5 1.0 1.5 2.00

1

2

3

4

5

6

0.0 0.5 1.0 1.5 2.0

0.10

0.08

0.06

0.04

0.02

0.00

0.0 0.5 1.0 1.5 2.0

30

28

26

24

22

20

18m 1.9max

1.21.3

1.5

1.8

1.71.6

1.4

1.1

1.1

1.1

(°) a

z s

e

Fig. 5. Loading–unloading curves between β, zs, κ and the bending moment for different values of mmax. The relative thickness is fixed at b=a¼ 0:1.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–88 81

effect of μ on rotations β and shifts zs of the neutral axis andcurvatures κ of the beam for the complete bending–unbendingprocess, when the beam is subjected to the maximum bendingmoment mmax ¼ 1:9, is shown. The cross-sectional model includedn¼22,400 square elements.

In the linear elastic case (μ¼ 1:0) rotation and shift remainconstant during the loading–unloading process, i.e. β6�63:7781and zs=a¼ 0:0, respectively, whereas curvature of the beam isproportional to the parameter m, i.e. κ=κe ¼m. The stress states forthe fully loaded and unloaded case, with the material parameterμ¼ 0:01 and maximum bending moment mmax ¼ 2:0, are plottedin Fig. 13.

Similar to the first numerical example, the contours, whichrepresent the stresses distribution over the cross-section, are nomore parallel to the neutral axis after unloading, see Fig. 13.

3.3. Rectangular L-beam

Consider a forming application (loadingþunloading) whererectangular L-beam, 10�15�1.5 mm in cross-section, is subjectedto the bending moment M(t), as shown in Fig. 14. The toolcomprises a fixed circular plate of radius R with a groove at angleφ and a handle with a small wheel, which fits perfectly to thegroove. As the beam is mounted to the tool at one end, the turn ofthe handle deforms it according to the radius of the plate. Afterun-mounting and springback of the beam, the final shape withradius of curvature 4R is obtained. We would like to find an angleφ at which a perfectly planar final shape of the beam is obtained.

The idea for solution arises from the observation of βðmÞ curves(e.g. the left diagram in Fig. 5). We can assume for beams ofasymmetric cross-section that during the elastic–plastic bending

a

n.a.

a

C

M t( )

yC

zC

by

z

b0

0

E

Et

Fig. 9. T-beam (left) and the stress–strain relations of material obeying the elastic-linear hardening rheological model (right).

1.2 1.4 1.6 1.8 2.0

0

2

4

6

8

10

12

efin

1.0 1.2 1.4 1.6 1.8 2.0

0.05

0.06

0.07

0.08

0.09

0.10

1.0 1.2 1.4 1.6 1.8 2.0

77.0

76.5

76.0

75.5

75.0

74.5

74.0

0.01

0.1

0.50.8

0.01

0.1

0.5

0.8

0.01

0.1

0.5

0.8

(°)

fin fina

z s

mmax mmax mmax

Fig. 10. Final rotations βfin, shifts zfins and curvatures κfin for different values of the material parameter μ.

mmax

1.0 1.2 1.4 1.6 1.8 2.0

2.0

1.8

1.6

1.4

1.2

1.0

1.0 1.2 1.4 1.6 1.8 2.00.00

0.02

0.04

0.06

0.08

0.10

1.0 1.2 1.4 1.6 1.8 2.012

10

8

6

4

2

0.01

0.1

0.5

0.8

mmaxmmax

(°)

s sa

z s

es

0.01

0.5

0.8

2.01.951.92.0

1.95

1.9

0.5

0.01

0.1

0.8

0.1

Fig. 11. Springback rotations βs, shifts zss and curvatures κs for different values of the material parameter μ.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8882

process there always exists a non-zero rotation of the neutral axis β,irrespective of the (constant) direction α of the bending momentM(t).The forming tool (with arbitrary φ) constrains the deformation of thebeam into a plane. But according to the previous assumption, theconstrained beam is also deformed in torsion. To obtain a perfectlyplanar shape of the beam after un-mounting, the torsional deforma-tion would have to vanish. Let us assume further that the vanishing ofthe torsional deformation occurs when the mean value of rotation ofthe neutral axis at the beginning β0 and that at the end of the loadingprocess βmax is equal to the direction α (the direction of the bendingmoment, cf. Figs. 14 and 15):

α¼ β0þβmax

2: ð16Þ

Here the rotation of the neutral axis βmax is the rotation β at themaximum bending moment Mmax, which is determined from thepredefined radius of the circular plate R, rotation βðMmaxÞ, shift of

0.0 0.5 1.0 1.5 2.00

1

2

3

4

5

6

0.0 0.5 1.0 1.5 2.00.00

0.02

0.04

0.06

0.08

0.10

0.0 0.5 1.0 1.5 2.0

76

74

72

70

68

66

64

mmm

(°) a

z s

e

0.01

0.1

0.5 0.8

0.1

0.50.8

0.01

0.1

0.50.8

0.01

mmax mmax

mmax

Fig. 12. Loading–unloading curves between β, zs, κ and the bending moment for different values of the material parameter μ (mmax ¼ 1:9).

0 0.2 0.4 0.80.6 1.00

0.2

0.4

0.6

0.8

1.0

y a

az

1.06703

0.44593

0.756480.91176

0.60121

0.01990

0.290660.13538

0.175170.330450.485720.641000.796270.951551.10682

Bending stress 0

m mmax

0

0.2

0.4

0.6

0.8

1.0

az

0 0.2 0.4 0.80.6 1.0y a

0.89881

0.35536

0.627080.76294

0.49122

0.05223

0.219490.08363

0.188090.323960.459820.595680.731540.867401.00327

Bending stress 0

m 0.0

Fig. 13. Fully loaded (left) and unloaded (right) stress states for μ¼ 0:01 and mmax ¼ 2:0.

R

circular plate beam

wheel

handle

Rbeam

wheel

handle

circular plate

yM t( )

z

Fig. 14. Scheme of the bending tool and produced bending moment during the forming process.

loadingprocess

M t( )

max

0

Mmax0

elastic plastic

Fig. 15. Vanishing of the torsional deformations during the loading process.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–88 83

the neutral axis zsðMmaxÞ and curvature of the beam κðMmaxÞ,Fig. 16. According to Mmax, the following expression has to besatisfied:

R¼ rP ¼ r�zP ¼ r�ðzPC�zsÞ: ð17Þ

Angle φ can be found in four steps:

1. Assume the direction α of the bending moment (see Fig. 14).2. Determine the maximum bending moment Mmax to satisfy the

condition (17).

3. Check the condition for vanishing of the torsional deforma-tions, Eq. (16). If the condition is satisfied, search for α isfinished, otherwise go to step 1.

4. Calculate φ from �αþφ¼ 901 (see Figs. 14 and 16).

If we consider material parameters of aluminum alloy 6060 T6(material properties are listed in Appendix A), the followingresults are obtained: α¼ �70:624441, φ¼ 19:375561, Mmax ¼12;159:14021 N mm, β0 ¼ �67:206651, βmax ¼ �74:042241,zsðMmaxÞ ¼ �0:38904 mm, κðMmaxÞ ¼ �9:66542� 10�3 mm�1-

rðMmaxÞ ¼ �103:46160 mm, rPðMmaxÞ ¼ �100:0 mm, zfins ¼�0:43965 mm, κfin ¼ �8:57985� 10�3 mm�1-rfin ¼�116:55215 mm, rP;fin ¼ �113:20583mm. Note that the cross-section is modeled by n¼14,100 square elements of dimensions0.05�0.05 mm. Radius of the plate is R¼ 100:0 mm, which meansthat according to the presented formulation the radius of curva-ture of the circular plate is �100.0 mm. Note that maximum straindid not exceed ε¼ 0:06346 (fiber N, Fig. 16).

Fig. 17 shows rotations β and shifts zs of the neutral axisand curvatures κ of the beam for the complete loading–unloadingprocess. The stress and strain states for the fully loaded case areplotted in Fig. 18, whereas the stress state after unloading is shownin Fig. 19. Contrary to the stress distribution, the contoursrepresenting the strain distribution over the cross-section arealways parallel to the neutral axis, Eq. (2).

In the case of kinematic hardening model, which is presentedin Appendix B, the final radius of curvature of fiber P is rP;fin ¼�113:22151mm. Obviously strain path reversal can be found insome fibers. The relative difference between results obtained byboth hardening models is only �0.01385%.

z

y

M

n. a.

R

r

P

z P

zCP

maxzs

zC

C yC

zzC

yC

y

circular

plate

N

z N

Fig. 16. Maximum bending moment Mmax which causes a given fiber of the cross-section (e.g. fiber P) to take a predefined radius of curvature R.

0 2 4 6 8 10 120.010

0.008

0.006

0.004

0.002

0.000

0 2 4 6 8 10 12

0.4

0.3

0.2

0.1

0.0

0 2 4 6 8 10 12

74

72

70

68

(°)

z(m

m)

s

) m

m(unloadingloading

unloadingloading

unloadingloading

M (Nm) M (Nm) M (Nm)

Fig. 17. Loading–unloading curves between β, zs, κ and the bending moment.

0 2 4 6 8 100

2

4

6

8

10

12

14

y (mm)

z (m

m)

Bending stress

211.413

92.709

152.061181.737

122.385

3.682

63.03433.358

25.99455.67085.346115.021144.697174.373204.049

(Mpa)

M t M( ) max

y (mm)

z (m

m)

Strain ( )

0.06346

0.03577

0.049610.05654

0.04269

0.01500

0.028850.02192

0.005770.012690.019610.026540.03346

M t M( ) max

0 2 4 6 8 100

2

4

6

8

10

12

14

Fig. 18. Fully loaded stress (left) and strain (right) states.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8884

4. Experiment

A custom-made forming tool was designed and manufacturedin-house to experimentally evaluate the proposed solution proce-dure and results obtained in numerical example 3.3. The sche-matics and its practical realization are shown in Figs. 14 and 20.

Note that, in general, this forming tool does not enable onlypure bending conditions. But, for this particular purpose (for acarefully prescribed φ), it enables the same final deformations ofthe beam as obtained from pure bending – the result is a circulararch. The circular plate of radius R¼ 100 mm with a groove atangle φ¼ 19:381 was used.

We performed experiments on 10 specimens, rectangularL-beams (10�15�1.5 mm in cross-section and L¼ 500 mm inlength) made of aluminum alloy 6060 T6. After the forming

process, the beams take a shape of a circular arch (essentially flat)as shown in Fig. 21. Just for illustration, a photo is shown inFig. 22 of spatial deformation of the beam caused by additionaltorsion which originated from the incorrect angle of the grooveφ¼ 23:551.

For each formed beam the final radius of curvature of fiber Pwas measured (see Fig. 16). The relative differences ϵr betweentheoretically (rP;fin ¼ �113:20583 mm) and experimentally deter-mined final radii of curvature of fiber P can be found in Table 1,where the average relative difference is �2.21770.699%(the uncertainty is the s.d. of the mean).

4.1. The effect of pre-strain on elastic modulus

In the present formulation and numerical examples it isassumed that the elastic modulus remains constant during theloading–unloading process. From additional experimental tests oncyclic loading of our aluminum specimens we observe an effect ofpre-strain on elastic modulus (see Figs. 23, 24 and Table 2).

The evolution of the elastic modulus can be described with ascalar function of change of the plastic strain response over thecomplete loading history εfp using a saturated type function whichhas been widely used for metallic materials [12,14,19,20]. Follow-ing the formalism from the literature, we model the elasticmodulus as

EðεfpÞ ¼ E0�ðE0�EsatÞð1�eð�ξεfpÞÞ; ð18Þwhere Esat and ξ are two material parameters, Esat is the saturatedelastic modulus and E0 is the initial elastic modulus. The initialvalue of elastic modulus of aluminum alloy 6060 T6 is E0 ¼67322:455 MPa. By fitting the measured elastic modulus duringthe unloading and using Eq. (18), we found Esat ¼ 49;807:318 MPaand ξ¼ 34:374. In Fig. 24, we show values of the elastic modulus(during unloading) from measurements and diagram of a fittingfunction at different pre-strains. It can be seen that the elastic

0 2 4 6 8 100

2

4

6

8

10

12

14Bending stress

190.782

81.742

136.262163.522

109.002

0.039

54.48127.221

27.30054.56081.820109.080136.340163.600

(Mpa)

y (mm)

z (m

m)

M t( ) 0.0 Nm

0.02871

0.05748

P

N

Fig. 19. The stress state after unloading. Strains in P and N.

Fig. 20. Custom-made forming tool.

Fig. 21. Plane deformation state of the formed beam (for angle φ¼ 19:381).

Fig. 22. Spatial deformation state of the formed beam (for angle φ¼ 23:551).

Table 1The comparison between theoretically and experimen-tally determined final radii of curvature of fiber P.

# rP;finexp ð mmÞ ϵr ð%Þ

1 �116.105 �2.5612 �114.982 �1.5693 �114.464 �1.1114 �115.980 �2.4515 �115.427 �1.9626 �114.934 �1.5277 �117.146 �3.4818 �115.319 �1.8679 �116.167 �2.61610 �116.630 �3.025Aver. �115.715 �2.217

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–88 85

modulus rapidly decreases and saturates to Esat as the pre-strainincreases.

This effect can easily be implemented in algorithm for updatingstress–strain relations presented in Section 2.1, where the stress–strain relation in the elastic region f e, Eq. (12), is now written as

f eðεmodj Þ ¼ signðεmod

j Þjεmodj jEj; ð19Þ

for jAf1;2;…;nNg, and some changes in steps 7 and 8 of thepseudo-code are implemented, i.e.

7. Elastic stress state� Calculate shift of the modified stress–strain relation:σsj ¼ �σj; εsj ¼ �ðΔεj�εsj Þ

� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞðjΔεjj�jΔσj=Eð0ÞjÞ

� Exit8. Elastic–plastic stress state

� Update yield stress: σ0; upj ¼ jσjj

� Calculate change of the plastic strain: Δεpj ¼ jΔεjj�jΔσj=Eð0Þj

� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞΔεpj

� Calculate change of the plastic response for a given incre-ment of the load: Δσfp

j ¼ σ0; upj �σ0

j ; Δεfpj ¼ jΔεjj�jΔσj=EjjþΔσfp

j =Ej� Calculate change of the plastic response over the completeloading history: σfp

j ¼ σfpj þΔσfp

j ; εfpj ¼ εfpj þΔεfpj

� Update elastic modulus: Ej ¼ Eðεfpj Þ� Update yield strain: ε0; upj ¼ σ0; up

j =Ej� Calculate shift of the modified stress–strain relation:σsj ¼ �signðσjÞσ0; up

j ; εsj ¼ �signðσjÞε0; upj� Exit

Finally, if the effect of pre-strain on elastic modulus is takeninto consideration and if the beam is curved along the circularplate of radius R¼ 100:0 mm as in numerical example 3.3, thefollowing results are obtained: zfins ¼ �0:44116 mm, κfin ¼�8:38478� 10�3 mm�1-rfin ¼ �119:26376 mm,rP;fin ¼ �115:92016 mm. The stress state after the unloading pro-cess is plotted in Fig. 25. In comparison to the results of the finalradii of curvature of fiber P determined by experiments, Table 1,the average relative difference ϵr ¼ 0:17770:683% is obtained.

As expected, the stresses in Fig. 25 are lower in comparisonto the stresses in Fig. 19, where the effect of pre-strain onelastic modulus is not considered. By decreasing of the elasticmodulus the formed beam exhibits more springback, i.e. rP;s ¼rP;fin�rPðMmaxÞ ¼ �15:92016 mm (in the case of E¼ const:, rP;s ¼�13:20583 mm).

5. Conclusion

We analyzed the problem of elastic–plastic bending andspringback of beams of asymmetric cross-section. We introduceda procedure for numerical computation of stress and strain statesof this geometric and material nonlinear problem. A completeloading history was taken into account including the effect ofthe local loading during the monotonic decrease of the load.The strains are described as a function of rotation and shift ofthe neutral axis and the curvature of the beam. The presentedalgorithm involves a scheme for updating stress–strain relations ofeach fiber of the cross-section, and is used to calculate the changeof the stress for each increment of the load. Numerical examplesconfirm a strong influence of maximum bending moment onthe final and springback rotation of the neutral axis, its shift,and curvature of the beam for different cross-sections and materi-als. Both isotropic and kinematic strain-hardening models are

0.00 0.02 0.04 0.06 0.08 0.10 0.120

50

100

150

200

6% 8%1% 2% 4%

engineering strain

)aPM(

sserts ginreenigne

Fig. 23. Experimental stress–strain curves of cyclic loading of the specimens madeof aluminum alloy 6060 T6 for five different pre-strains.

0.00 0.02 0.04 0.06 0.08

50 000

55 000

60 000

65 000

70 000

experimentfitted

0

fp

)aPM(

suludom citsale

E

pre-strain

Fig. 24. Elastic modulus as a function of pre-strain.

Table 2The effect of pre-strain on elastic modulus. The linear chord modulus model wasadopted.

# ε ð%Þ E ðMPaÞ

1 ε0 ¼ 0:2766 67,322.4552 1.0 62,343.2723 2.0 59,158.4124 4.0 55,556.0245 6.0 52,865.8556 8.0 50,219.493

0 2 4 6 8 100

2

4

6

8

10

12

14

y (mm)

z (m

m)

Bending stress

189.725

82.225

135.975162.850

109.100

1.600

55.35028.475

25.27552.14979.024105.899132.774159.649

(Mpa)

M t( ) 0.0 Nm

0.02804

0.05620

P

N

Fig. 25. The stress state after unloading. The effect of pre-strain on elastic modulusis taken into consideration. Strains in P and N.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8886

considered. It was shown that for the chosen example the relativedifference between results obtained by both hardening modelsis practically negligible, but not zero (the results lie within0.01385%). Note that our solution procedure can be used forarbitrary loading regime (such as multi-cycle and non-monotonicloading) and also more complex materials (composite, functionallygraded, etc.).

A custom-made forming tool was designed and manufacturedin-house to experimentally evaluate the proposed solution proce-dure. A rectangular L-beam made of an aluminum alloy 6060 T6was used and its material properties were carefully measured.The first task in this example was to determine an appropriateangle φ of the groove on circular plate at which a perfectly planarfinal shape of the beam would be obtained after forming(to remove the effect of torsion). It was shown that relativedifference between theoretically predicted and experimental mea-surements of the final radius of curvature of the formed beam was

within �2,21770,699%, if the elastic modulus was considered toremain constant during the deformation. Including an additionalfeature, the effect of pre-strain on elastic modulus (evolution ofthe elastic modulus), in our algorithm, enabled a better descriptionof a real material behavior. We obtained an excellent agreementwith experiments. The results differed from experiments by onlyϵr ¼ 0:17770:683% in average.

Appendix A. Measuring mechanical properties of aluminumalloy 6060 T6

Material properties were measured on a Zwick Z050 measuringdevice, equipped with Multisens extensometers. The measuredstress–strain curves of nine flat specimens (cut from the 10�15�1.5 mm L-beams) are illustrated in Fig. 26. Nine tests (one had tobe discarded due to the technical difficulties) were performed atthe same experimental conditions (cross-head speed¼10 mm/min, cross-head travel resolution¼40:0 μm and temperature¼20 1C) and then piecewise approximated by seven linear functions,as shown in Fig. 26 and given Table 3.

Note that Et is approximated to be constant after 0.08 of strain,i.e. Et ¼ 24:636 MPa for ε40:08.

Appendix B. Algorithm for updating stress–strain relationswhen the kinematic hardening model is taken intoconsideration

A computational procedure for updating stress strain relationsσupj ðεupj Þ of j-th fiber of the cross-section prepared for the next

increment of the load, when the kinematic hardening model istaken into consideration, can be expressed in the way very similarto the isotropic hardening model, see Fig. 27. Updated stress–strain relation σup

j ðεupj Þ can be described by Eqs. (9), (10) and (11)and (11), where the yield point, expressed by σj0 and εj0, is constantfor the complete loading–unloading process.

Additional parameter σsylj is involved and represents a shift of

the upper and lower yield limits according to σj-εj coordinatesystem. Here, we describe the procedure for updating the stress–strain relations of j-th fiber in each increment of the load.Graphical description is given in Fig. 27 for the first increment ofthe load.

Pseudo-code.

1. Start with stored known variables for the previous load, i.e.at time t0�Δt: σj, εj, σfp

j , εfpj , σ

sj , ε

sj , ε

pj and σsyl

j2. An increment of the load causes change of the strain: Δεj

)aPM(

sserts ginreenigne

0.00 0.02 0.04 0.06 0.08 0.10 0.120

50

100

150

200

experimentapproximation

engineering strain

Fig. 26. Experimental and approximated stress–strain curves of aluminum alloy6060 T6.

Table 3Approximated stress–strain curve of aluminum alloy 6060 T6.

# σ ðMPaÞ ε ð Þ Et ðMPaÞ

1 0.000 0.0002 σ0 ¼ 186:187 ε0 ¼ 2:766� 10�3 E¼67,322.455

3 199.881 0.025 615.9104 204.809 0.035 492.7285 208.319 0.045 351.0696 210.475 0.055 215.5687 211.584 0.065 110.8648 211.953 0.080 24.636

0

0

yield limit

p

up. yield limit

j

j0j

j

fpj

j

0j

0j

j

fpj

0j

0j

j

j

1

pf ( ; , )0j

0j

ef ( )j

j

modjpf ( ; , , , )0

j0j

fpj

fpj

ef ( )modj

sylj

mod

mod

updated yield limit

j

j

0j

sj

sj

upj

upj

2

ef ( )modj

modjpf ( ; , , , )0

j0j

fpj

fpj

0j

0j

0j

Fig. 27. Evolution of the updated stress–strain relation σupj ðεupj Þ of the j-th fiber, jAf1;2;…;nNg. Kinematic hardening is considered.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–88 87

3. Calculate change of the stress state: Δσj ¼ σupj ðΔεjÞ

4. Calculate current stress and strain state: σj ¼ σjþΔσj;εj ¼ εjþΔεj

5. Check for the yield condition:� If jσj�σsyl

j j�σ0j o0; Then Goto step 6

� Else Goto step 76. Elastic stress state

� Calculate shift of the modified stress–strain relation:σsj ¼ �ðσj�σsyl

j Þ; εsj ¼ �ðΔεj�εsj Þ� Exit

7. Elastic–plastic stress state� Calculate change of the plastic strain: Δεpj ¼ jΔεjj�jΔσj=Ej� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞΔεpj� Calculate change of the plastic response for a given incre-ment of the load:○ If Δσj ¼ 0; Then Δσfp

j ¼ 0○ Else Δσfp

j ¼ jΔσjj�σ0j �signðΔσjÞσs

j

Δεfpj ¼Δεpj þΔσfpj =E

� Calculate change of the plastic response over the completeloading history: σfp

j ¼ σfpj þΔσfp

j ; εfpj ¼ εfpj þΔεfpj

� Calculate shift of the modified stress–strain relation:σsj ¼ �signðσj�σsyl

j Þσ0j ; ε

sj ¼ �signðσj�σsyl

j Þε0j� Calculate shift of the yield limits:

○ If Δσj ¼ 0; Then σsylj ¼ σsyl

j○ Else σsyl

j ¼ σj�signðΔσjÞσ0j

� Exit

If the effect of pre-strain on elastic modulus is taken intoconsideration, some changes in steps 6 and 7 of the pseudo-codeare implemented, i.e.

6. Elastic stress state� Calculate shift of the modified stress–strain relation:σsj ¼ �ðσj�σsyl

j Þ; εsj ¼ �ðΔεj�εsj Þ� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞðjΔεjj�jΔσj=Eð0ÞjÞ

� Exit7. Elastic–plastic stress state

� Calculate change of the plastic strain: Δεpj ¼ jΔεjj�jΔσj=Eð0Þj

� Calculate the total plastic strain: εpj ¼ εpj þsignðΔεjÞΔεpj� Calculate change of the plastic response for a given incre-ment of the load:○ If Δσj ¼ 0; Then Δσfp

j ¼ 0○ Else Δσfp

j ¼ jΔσjj�σ0j �signðΔσjÞσs

j

Δεfpj ¼ jΔεjj�jΔσj=EjjþΔσfpj =Ej

� Calculate change of the plastic response over the completeloading history: σfp

j ¼ σfpj þΔσfp

j ; εfpj ¼ εfpj þΔεfpj

� Update elastic modulus: Ej ¼ Eðεfpj Þ� Update yield strain: ε0j ¼ σ0

j =Ej� Calculate shift of the modified stress–strain relation:σsj ¼ �signðσj�σsyl

j Þσ0j ; ε

sj ¼ �signðσj�σsyl

j Þε0j

� Calculate shift of the yield limits:○ If Δσj ¼ 0; Then σsyl

j ¼ σsylj

○ Else σsylj ¼ σj�signðΔσjÞσ0

j� Exit

References

[1] Kosel F, Videnic T, Kosel T, Brojan M. Elasto-plastic springback of beamssubjected to repeated bending/unbending histories. J Mater Eng Perform2011;20:846–54.

[2] Yu TX, Johnson W. Influence of axial force on the elastic–plastic bending andspringback of a beam. J Mech Work Technol 1982;6:5–21.

[3] Johnson W, Yu TX. On springback after the pure bending of beams and platesof elastic work-hardening materials III. Int J Mech Sci 1981;23:687–95.

[4] Xu Y, Zhang LC, Yu TX. The elastic–plastic pure bending and springback ofL-shaped beams. Int J Mech Sci 1987;29:425–33.

[5] Baragetti S. A theoretical study on nonlinear bending of wires. Meccanica2006;41:443–58.

[6] Li KP, Carden WP, Wagoner RH. Simulation of springback. Int J Mech Sci2002;44:103–22.

[7] Wagoner RH, Li M. Simulation of springback: through-thickness integration.Int J Plast 2007;23:345–60.

[8] Kazan R, Firat M, Tiryaki AE. Prediction of springback in wipe-bending processof sheet metal using neural network. Mater Des 2009;30:418–23.

[9] Panthi SK, Ramakrishnan N, Pathak KK, Chouhan JS. An analysis of springbackin sheet metal bending using finite element method (FEM). J Mater ProcessTechnol 2007;186:120–4.

[10] Ragai I, Lazim D, Nemes JA. Anisotropy and springback in draw-bending ofstainless steel 410: experimental and numerical study. J Mater Process Technol2005;166:116–27.

[11] Vladimirov IN, Pietryga MP, Reese S. Prediction of springback in sheet formingby a new finite strain model with nonlinear kinematic and isotropic hard-ening. J Mater Process Technol 2009;209:4062–75.

[12] Ghaei A. Numerical simulation of springback using an extended returnmapping algorithm considering strain dependency of elastic modulus. Int JMech Sci 2012;65:38–47.

[13] Cleveland RM, Ghosh AK. Inelastic effects on springback in metals. Int J Plast2002;18:769–85.

[14] Zang S-L, Lee M-G, Kim JH. Evaluating the significance of hardening behaviourand unloading modulus under strain reversal in sheet springback prediction.Int J Mech Sci 2013;77:194–204.

[15] Vin LJ, Streppl AH, Singh UP, Kals HJJ. A process model for air bending. J MaterProcess Technol 1996;57:48–54.

[16] Yu HY. Variation of elastic modulus during plastic deformation and itsinfluence on springback. Mater Des 2009;30:846–50.

[17] Abdel-Karim M. Effect of elastic modulus variation during plastic deformationon uniaxial and multiaxial ratchetting simulations. Eur J Mech A/Solid2011;30:11–21.

[18] Sun L, Wagoner RH. Complex unloading behavior: nature of the deformationand its consistent constitutive representation. Int J Plast 2011;27:1126–44.

[19] Zang S-L, Lee M-G, Sun L, Kim JH. Measurement of the Bauschinger behavior ofsheet metals by three-point bending springback test with pre-strained strips.Int J Plast 2014;59:84–107.

[20] Lee J, Lee J-Y, Barlat F, Wagoner RH, Chung K, Lee M-G. Extension of quasi-plastic-elastic approach to incorporate complex plastic flow behavior—appli-cation to springback of advanced high-strength steels. Int J Plast2013;45:140–59.

[21] Brojan M, Sitar M, Kosel F. On static stability of nonlinearly elastic Euler'scolumns obeying the modified Ludwick's law. Int J Struct Stab Dyn2012;12:1250077(1)–19).

[22] Burden RL, Faires JD. Numerical analysis. 9th ed. Boston: Brooks/Cole; 2010.[23] Jung JH, Kang TJ. Large deflection analysis of fibers with nonlinear elastic

properties. Text Res J 2005;75:715–23.[24] Brojan M, Videnic T, Kosel F. Large deflections of nonlinearly elastic non-

prismatic cantilever beams made from materials obeying the generalizedLudwick constitutive law. Meccanica 2009;44:733–9.

M. Sitar et al. / International Journal of Mechanical Sciences 90 (2015) 77–8888