Efficient force prediction for incremental sheet forming and experimental validation

Abstract

Incremental sheet forming (ISF) has been attractive during the last decades because of its greater flexibility, increased formability and reduced forming forces. However, traditional finite element simulation used for force prediction is significantly time consuming. This study aims to provide an efficient analytical model for tangential force prediction. In the present work, forces during the cone-forming process with different wall angles and step-down sizes are recorded experimentally. Different force trends are identified and discussed with reference to different deformation mechanisms. An efficient model is proposed based on the energy method to study the deformation zone in a cone-forming process. The effects of deformation modes from shear, bending and stretching are taken into account separately by two sub-models. The final predicted tangential forces are compared with the experimental results which show an average error of 6 and 11 % in respect to the variation of step-down size and wall angle in the explored limits, respectively. The proposed model would greatly improve the prediction efficiency of forming force and benefit both the design and forming process.

Appendices

Appendix 1

1.1 Calculation of critical angles

As we can see from Fig. 16, the deformation zone which is being formed in current path can be divided into six regions. In radial direction, the deformation zone is divided into two parts by the radius of (r _i  + ∆r). In circumferential direction, three critical angles (θ _c ,  θ _t and  θ _t ) are used to define the assumed flow line. In Fig. 16, we can see that regions Ia, IIa and IIIa are undeformed areas during last pass so these are flat before the current forming pass, whereas regions Ib, IIb and IIIb are previously deformed in the last forming pass. Vertical positions for the sheet before the current forming pass are,

Fig. 16

Divided deformation zone in the cone-forming process

$$ {Z}_{sa}={Z}_t+\varDelta z\kern3em \left({r}_i\le \mathrm{r}\le {r}_i+\varDelta r\right) $$

(34)

$$ {Z}_{sb}={Z}_t+\varDelta z+{r}_t-\sqrt{{r_t}^2-{\left(r-{r}_i-\varDelta z \tan \left(90-\alpha \right)\right)}^2}\kern1.5em \left({r}_i+\varDelta r\le \mathrm{r}\le {r}_o\right) $$

(35)

a) Calculation of θ _c

As previously defined,  Z _c is the conjunct point for curve Zf _c and tool head surface, so θ _ca is calculated by solving the following two equations,

$$ \left\{\begin{array}{c}\hfill {Z}_c=M\left({Z}_{s1}-{Z}_o\right)+{Z}_o\hfill \\ {}\hfill {Z}_c={Z}_{t0}+{r}_t-\sqrt{{r_t}^2-{\left(r \cos \theta -{r}_i\right)}^2-{\left(r \sin \theta \right)}^2}\hfill \end{array}\right.\kern5.5em $$

(36)

According to the geometric relations of the tool head, we have,

$$ \left\{\begin{array}{c}\hfill {Z}_o={Z}_{t0}+\left({r}_t-{r}_{tc}\right)\hfill \\ {}\hfill {r}_{tc}=\sqrt{{r_t}^2-{\left(r-{r}_i\right)}^2}\hfill \end{array}\right.\kern3em $$

(37)

The contact angle θ _ca can be solved from Eqs. (37) and (38) gives

$$ {\theta}_{ca}={ \cos}^{-1}\frac{{\left(\sqrt{{r_t}^2-{\left(r-{r}_i\right)}^2}-M\left({Z}_{sa}-{Z}_o\right)\right)}^2-{r_t}^2+{r}^2+{r_i}^2}{2r{r}_i} $$

(38)

Similarly, the contact angle θ _cb which separates regions Ib and IIb can be solved as,

$$ {\theta}_{cb}={ \cos}^{-1}\frac{{\left(\sqrt{{r_t}^2-{\left(r-{r}_i\right)}^2}-M\left({Z}_{sb}-{Z}_o\right)\right)}^2-{r_t}^2+{r}^2+{r_i}^2}{2r{r}_i} $$

(39)

b) Calculation of θ _t

When  r _i  ≤ r ≤ r _i  + Δr, θ _ta is calculated to define regions IIa and IIIa,

$$ \left\{\begin{array}{c}\hfill {Z}_{ta}={Z}_{t0}+\varDelta z\hfill \\ {}\hfill {Z}_t={Z}_{t0}+{r}_t-\sqrt{{r_t}^2-{\left(r \cos \theta -{r}_i\right)}^2-{\left(r \sin \theta \right)}^2}\hfill \end{array}\right. $$

(40)

Let Z _ta equals  Z _t , the angle θ _ta can be solved as,

$$ {\theta}_{ta}={ \cos}^{-1}\frac{{\left({r}_t-\varDelta z\right)}^2-{r_t}^2+{r}^2+{r_i}^2}{2r{r}_i} $$

(41)

Similarly, for calculation of θ _tb , we have

$$ \left\{\begin{array}{c}\hfill {Z}_{tb}={Z}_{t0}+\varDelta z+{r}_t-\sqrt{{r_t}^2-{\left(r-{r}_i-\varDelta z \tan \left(90-\alpha \right)\right)}^2}\hfill \\ {}\hfill {Z}_t={Z}_{t0}+{r}_t-\sqrt{{r_t}^2-{\left(r \cos \theta -{r}_i\right)}^2-{\left(r \sin \theta \right)}^2}\hfill \end{array}\right. $$

(42)

$$ {\theta}_{tb}={ \cos}^{-1}\frac{{\left(\sqrt{{r_t}^2-{\left(r-{r}_i-{d}_z \tan \left(90-\alpha \right)\right)}^2}-\varDelta z\right)}^2-{r_t}^2+{r}^2+{r_i}^2}{2r{r}_i} $$

(43)

c) Calculation of θ _s

The deformation zone is assumed to be a circle with the radius of r _o  − r _i , so the following equations set can be given as,

$$ \left\{\begin{array}{c}\hfill {X}^2+{Y}^2={r}^2\hfill \\ {}\hfill {\left(X-{r}_i\right)}^2+{Y}^2={\left({r}_o-{r}_i\right)}^2\hfill \end{array}\right. $$

(44)

The angle θ _s can be solved as,

$$ {\theta}_s={ \sin}^{-1}\frac{Y}{r} $$

(45)

Appendix 2

2.1 Calculation of average yield strength and equivalent strain

For each of the six regions, do the integral for equivalent strain. Take region Ia for example,

$$ {\overline{\varepsilon}}_{I\alpha}={\displaystyle {\iint}_0^{\theta_{ca}}}\frac{2}{\sqrt{3}}\left({\varepsilon_{\theta z}}^2+{\varepsilon_{rz}}^2\right)\mathrm{d}\theta \mathrm{d}r $$

(46)

On the velocity discontinuity surface,

$$ {\overline{\varepsilon}}_{disa}=\frac{2}{\sqrt{3}}{\varepsilon}_{\theta z}=\frac{1}{\sqrt{3}}\frac{\left|\varDelta v\right|}{v_{\theta }}.\kern1.75em $$

(47)

The average equivalent strain is given as,

$$ {\overline{\varepsilon}}_{\mathrm{avg}}=\frac{{\overline{\varepsilon}}_{\mathrm{total}}}{r_o-{r}_i}=\frac{{\overline{\varepsilon}}_{\mathrm{Ia}}+{\overline{\varepsilon}}_{\mathrm{Ib}}+{\overline{\varepsilon}}_{\mathrm{II}\upalpha}+{\overline{\varepsilon}}_{\mathrm{II}\mathrm{b}}+{\overline{\varepsilon}}_{\mathrm{II}\mathrm{I}\upalpha}+{\overline{\varepsilon}}_{\mathrm{II}\mathrm{I}\mathrm{b}}+{\overline{\varepsilon}}_{disa}+{\overline{\varepsilon}}_{disb}}{r_o-{r}_i} $$

(48)

By assuming the swift type work hardening law σ _eq  = K(ε ₀ + ε)ⁿ for a sheet metal, the average yield strength is obtained as,

$$ {y}_0=\frac{{\displaystyle {\int}_0^{{\overline{\varepsilon}}_{\mathrm{avg}}}}{\sigma}_{eq}\mathrm{d}\varepsilon }{{\displaystyle {\int}_0^{{\overline{\varepsilon}}_{\mathrm{avg}}}}\mathrm{d}\varepsilon }, $$

(49)

where ε is the plastic strain. Also, K, n and ε ₀ are the material parameters.