Study on springback from thermal-mechanical boundary condition imposed to V-bending and L-bending processes coupled with infrared rays local heating

Abstract

This paper shows differences of thermal-mechanical boundary condition and springback between V-bending and L-bending processes. Although a recently conducted study showed that an infrared (IR) local heating method substantially reduces springback of dual-phase (DP) 980 steel sheet in V-bending, its application to L-bending, which is one of widely used forming processes in industrial applications, has not been sufficiently reported. In the L-bending experiment conducted in this work, springback does not sufficiently decrease as much as it does in the V-bending test, even though the sheet is deformed under the same temperature condition. For analysis of the difference in springback, thermo-elastic-plastic simulation for both V-bending and L-bending processes were conducted. The numerical analysis shows that the V-bending and L-banding processes have significantly different thermal and mechanical boundary conditions even though both processes go through a bending deformation. The differences in the boundary conditions have a strong effect on the thermal-mechanical behavior of the blank, so that the springback results are different between the two bending processes. Finally, it is also shown that the L-bending process requires a higher temperature condition than the V-bending process in order to sufficiently reduce springback.

Appendix

Stress integration

The incremental strain-stress relation in the thermal-mechanical deformation is given by

$$ d\boldsymbol{\sigma} =\boldsymbol{C}d{\boldsymbol{\varepsilon}}^e+d\boldsymbol{C}{\boldsymbol{\varepsilon}}^e $$

(A.1)

where d ε ^e = d ε − d ε ^th − dε ^p and $d\boldsymbol{C}=\left(\frac{d\boldsymbol{C}}{dT}\right) dT$.

d σ means the increment of stress, d ε is the increment of total strain, d ε ^e is the increment of elastic strain, d ε ^th is the increment of thermal strain. d ε ^p is the increment of plastic strain, and d C means the increment of mechanical modulus due to the temperature change, dT. In the numerical implementation, eq. (A.1) is accumulated in a step-wise manner as below:

$$ {\boldsymbol{\sigma}}_{n+1}={\boldsymbol{\sigma}}_n+d{\boldsymbol{\sigma}}_{n+1}={\boldsymbol{\sigma}}^T-\left(\boldsymbol{C}+d\boldsymbol{C}\right)d{\boldsymbol{\varepsilon}}_{n+1}^p $$

(A.2)

where ${\boldsymbol{\sigma}}^T=\left[{\boldsymbol{\sigma}}_n+\boldsymbol{C}\left(d{\boldsymbol{\varepsilon}}_{n+1}-d{\boldsymbol{\varepsilon}}_{n+1}^{th}\right)+d\boldsymbol{C}\left({\boldsymbol{\varepsilon}}_{n+1}-{\boldsymbol{\varepsilon}}_{n+1}^{th}-{\boldsymbol{\varepsilon}}_n^p\right)\right]$.

σ ^T is the trial stress, and n is the step number. The increment of plastic strain ($d{\boldsymbol{\varepsilon}}_{n+\mathtt{1}}^p$) is described in the associated rule, as below:

$$ d{\boldsymbol{\varepsilon}}_{\mathtt{n}+\mathtt{1}}^p={\left(\frac{\partial \overline{\sigma}}{\partial \boldsymbol{\sigma}}\right)}_{\mathtt{n}+\mathtt{1}}d{\overline{\varepsilon}}_{\mathtt{n}+\mathtt{1}}^{\mathtt{p}}\left(={\hat{\boldsymbol{e}}}_{\mathtt{n}+\mathtt{1}}{\gamma}_{\mathtt{n}+\mathtt{1}}\right) $$

(A.3)

where $\hat{\boldsymbol{e}}=\frac{\partial \overline{\sigma}}{\partial \boldsymbol{\sigma}},\kern0.5em {\gamma}_{\mathtt{n}+\mathtt{1}}=d{\overline{\varepsilon}}_{\mathtt{n}+\mathtt{1}}^{\mathtt{p}}$.

where $\overline{\sigma}$ is the effective stress, and $d{\overline{\varepsilon}}^{\mathtt{p}}$is the increment of effective plastic strain in the associated rule. In order to determine the value of $d{\overline{\varepsilon}}_{n+1}^p$, the following three functions (A.4 – A.6) are employed based on the procedure proposed by Yoon et al. [33]:

$$ {g}_1\left({\gamma}_{\mathtt{n}+\mathtt{1}}\right)=\overline{\sigma}\left({\boldsymbol{\sigma}}_{n+1}\right)-{\rho}_{n+1}\left({\overline{\varepsilon}}_{\mathtt{n}}^p+{\gamma}_{n+1}\right)=0 $$

(A.4)

$$ {\boldsymbol{g}}_2\left({\gamma}_{n+1}\right)={\boldsymbol{\sigma}}_{n+1}-{\boldsymbol{\sigma}}^T+\left(\boldsymbol{C}+d\boldsymbol{C}\right){\hat{\boldsymbol{e}}}_{n+1}{\gamma}_{n+1}=0 $$

(A.5)

$$ {g}_3\left({\gamma}_{n+1}\right)=\left({\rho}_{n+1}-{\rho}_n\right){\mathtt{H}}^{-1}-{\gamma}_{n+1}=0 $$

(A.6)

where ρ is the yield stress, and H is the discrete slope in the work-hardening curve. Since above functions (A.4 – A.6) are nonlinear equations, a linearization process is needed for them. After the linearization, the following relationship is obtained:

$$ d{\gamma}_{n+1}^i=\frac{g_1\left({\gamma}_{\mathtt{n}+\mathtt{1}}^{\mathtt{i}}\right)-{\widehat{\boldsymbol{e}}}_{n+1}^i{\left({\boldsymbol{E}}_{n+1}^i\right)}^{-1}{\boldsymbol{g}}_{\boldsymbol{2}}\left({\gamma}_{\mathtt{n}+\mathtt{1}}^{\mathtt{i}}\right)+{g}_3\left({\gamma}_{\mathtt{n}+\mathtt{1}}^{\mathtt{i}}\right)H}{H+{\widehat{\boldsymbol{e}}}_{\mathtt{n}+\mathtt{1}}^i{\left({\boldsymbol{E}}_{n+1}^i\right)}^{-1}{\widehat{\boldsymbol{e}}}_{n+1}^i+{\widehat{\boldsymbol{e}}}_{n+1}^i{\left({\boldsymbol{E}}_{n+1}^i\right)}^{-1}{\boldsymbol{C}}^{-1}d\boldsymbol{C}{\widehat{\boldsymbol{e}}}_{n+1}^i} $$

(A.7)

where ${\boldsymbol{E}}_{n+1}^i=\left[{\boldsymbol{C}}^{-1}+{\left(\frac{\partial \hat{\boldsymbol{e}}}{\partial \boldsymbol{\sigma}}\right)}_{n+1}^i{\gamma}_{n+1}^i+{\boldsymbol{C}}^{-1}d\boldsymbol{C}{\gamma}_{n+1}^i{\left(\frac{\partial \hat{\boldsymbol{e}}}{\partial \boldsymbol{\sigma}}\right)}_{n+1}^i\right]$,

where i is the iteration number. Then, $d{\overline{\varepsilon}}_{n+1}^p$ is given by:

$$ d{\overline{\varepsilon}}_{n+1}^p=\sum d{\gamma}_{n+1}^i $$

(A.8)

$d{\boldsymbol{\varepsilon}}_{n+1}^p$ can be obtained by substituting eq. (A.8) into eq. (A.3), then, the stress (σ _n + 1) is obtained by substituting eq. (A.3) into eq. (A.2).

Consistent tangent modulus

Taking the consistency condition of eq. (A.4) leads to

$$ \hat{\boldsymbol{e}}\cdot d\boldsymbol{\sigma} -{h}^{\hbox{'}} d\gamma =0 $$

(A.9)

where h ^’ ($\equiv \frac{\boldsymbol{d}\boldsymbol{\rho }}{\boldsymbol{d}\boldsymbol{\gamma }}$) means the instantaneous slope. The following relation (A.10) is needed to get the consistent tangent modulus.

$$ d\boldsymbol{\sigma} =\left[{\boldsymbol{C}}^{-1}+\gamma \frac{\partial \hat{\boldsymbol{e}}}{\partial \boldsymbol{\sigma}}+\left({\boldsymbol{C}}^{-1}d\boldsymbol{C}\right)\gamma \frac{\partial \hat{\boldsymbol{e}}}{\partial \boldsymbol{\sigma}}\right]\left[d\boldsymbol{\varepsilon} - d\gamma \hat{\boldsymbol{e}}+\left({\boldsymbol{C}}^{-1}d\boldsymbol{C}\right)d\boldsymbol{\varepsilon} -\left({\boldsymbol{C}}^{-1}d\boldsymbol{C}\right) d\gamma \hat{\boldsymbol{e}}\right] $$

(A.10)

Substituting eq. (A.10) into eq. (A.9) provides below equation:

$$ d\gamma =\frac{\hat{\boldsymbol{e}}\overline{\boldsymbol{C}}d\boldsymbol{\varepsilon} +\hat{\boldsymbol{e}}\overline{\boldsymbol{C}}\left({\boldsymbol{C}}^{-1}\right)d\boldsymbol{C}d\boldsymbol{\varepsilon}}{h^{\hbox{'}}+\hat{\boldsymbol{e}}\overline{\boldsymbol{C}}\hat{\boldsymbol{e}}+\hat{\boldsymbol{e}}\overline{\boldsymbol{C}}\left({\boldsymbol{C}}^{-1}\right)d\boldsymbol{C}\hat{\boldsymbol{e}}} $$

(A.11)

Then, the consistent tangent modulus (${\boldsymbol{C}}^{\mathtt{alg}}$) can be derived from substituting eq. (A.11) into eq. (A.10):

$$ {\boldsymbol{C}}^{a\lg }=\left[\overline{\boldsymbol{C}}\left(\boldsymbol{I}+{\boldsymbol{C}}^{-\boldsymbol{1}}d\boldsymbol{C}\right)-\frac{\left[\overline{\boldsymbol{C}}\widehat{\boldsymbol{e}}+\overline{\boldsymbol{C}}\left({\boldsymbol{C}}^{-\boldsymbol{1}}\right)\boldsymbol{d}C\widehat{e}\right]\otimes \left[\overline{\boldsymbol{C}}\widehat{\boldsymbol{e}}+\overline{\boldsymbol{C}}\left({\boldsymbol{C}}^{-\boldsymbol{1}}\right)d\boldsymbol{C}\widehat{\boldsymbol{e}}\right]}{h^{\boldsymbol{\hbox{'}}}+\widehat{\boldsymbol{e}}\overline{\boldsymbol{C}}\widehat{\boldsymbol{e}}+\widehat{\boldsymbol{e}}\overline{\boldsymbol{C}}\left({\boldsymbol{C}}^{-\boldsymbol{1}}\right)d\boldsymbol{C}\widehat{\boldsymbol{e}}}\right] $$

(A.12)