Stress State Analysis of Radial Stress Superposed Bending

Abstract

Radial stress superposed bending is a sheet metal bending process, which superposes predetermined radial stresses. Stress superposition is mandatory to enable the reduction of the triaxiality in bending, resulting in delayed damage evolution and an improved product performance. The knowledge of the stress state is essential for damage-controlled bending as the triaxiality is the driving force for the void evolution. To control the stress state in radial stress superposed bending, an additional counter force responsible for the pressure in the outer fiber is applied. To predict the effect of the counter force on the radial stress and the triaxiality an analytical model is proposed. The prediction of the reaction forces in the system is required for the process design and for the calculation of the stress superposition. The stress state for plane strain bending with stress superposition is derived, and pressure calculations are made using the theory of Hertz. The model and the assumptions are verified in numerical and experimental studies for various counter pressures and bending ratios. Finally, a discussion of the load path depending on the transient counter pressure is carried out and experimental evidence for a inhibited damage evolution due to stress superposition is given.

Appendices

Appendix 1

1.1 Friction Coefficient Declaration

• µ₁ Friction between lower bearing sheet and rotating tool

• µ₂ Friction between sheet and rotating tool

• µ₃ Friction between sheet and die

• µ₄ Friction between upper tools and sheet

• µ₅ Friction between upper tools and shell

Appendix 2

2.1 Determination of the Bending Angle

Determination of applied distances:

$$s_{6} = r_{\text{g}} + s_{3} + t + r_{\text{p}}$$

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$$s_{7} = h - s_{6}$$

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$$s_{8} = g/2 + r_{g}$$

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$$s_{9} = \sqrt {s_{7}^{2} + s_{8}^{2} }$$

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$$s_{10} = \sqrt {s_{6}^{2} - s_{9}^{2} }$$

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Determination of resulting angles:

$$\beta_{1} = \arctan \,\,\,\,\,\left( {\frac{{s_{6} }}{{s_{9} }}} \right)$$

(26)

$$\beta_{2} = \arctan \,\,\,\,\,\left( {\frac{{s_{7} }}{{s_{8} }}} \right)$$

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$$\alpha = 2 \cdot (\beta_{1} + \beta_{2} )$$

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Appendix 3

3.1 Upper Rotating Tools

The equilibrium of moments at the apparent punch center point C of one upper rotating parts (Fig. 10) leads to equation:

$$(N_{\text{l}} + N_{\text{p}} ) \cdot r_{\text{b}} \cdot \mu_{4} + N_{\text{ps}} \cdot r_{\text{up}} \cdot \mu_{5} - N_{\text{l}} \cdot (s_{4} - s_{5} ) = 0$$

(29)

The two rotating upper parts can be considered as one FBD for the equilibrium of vertical forces as they are connected in a cylindrical bearing (Fig. 10). This leads to equation:

$$\begin{aligned} 2 \cdot (N_{\text{l}} + N_{\text{ps}} ) \cdot (\cos (\alpha /2) - \sin (\alpha /2) \cdot \mu_{4} ) \\ & \quad - 2 \cdot N_{\text{ps}} \cdot (\cos (\varphi_{2} ) + \sin (\varphi_{2} ) \cdot \mu_{5} = 0 \\ \end{aligned}$$

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3.2 Lower Rotating Tools

The sum of the vertical forces leads to Eq. 31 and the sum of horizontal forces to Eq. 32

$$\begin{aligned} & N_{\text{b1}} + N_{\text{b2}} + N_{\text{r}} - N_{\text{d}} \\ & \quad - N_{\text{bs}} \cdot \cos (\varphi_{2} ) + T_{\text{bs}} \cdot \sin (\varphi_{2} ) = 0 \\ \end{aligned}$$

(31)

$$\begin{aligned} & T_{\text{b1}} + T_{\text{b2}} + T_{\text{r}} - T_{\text{d}} \\ & \quad - T_{\text{bs}} \cdot \cos (\varphi_{1} ) - N_{\text{bs}} \cdot \sin (\varphi_{1} ) = 0 \\ \end{aligned}$$

(32)

The equation for the equilibrium of moments around the apparent punch center point is given in equation:

$$\begin{aligned} & N_{\text{b1}} \cdot s_{4} + T_{\text{b1}} \cdot (t + r_{\text{p}} ) + N_{\text{b2}} \cdot s_{2} + T_{\text{b2}} \cdot (t + r_{\text{p}} ) \\ & \quad + N_{\text{r}} \cdot s_{1} + T_{\text{r}} \cdot (t + r_{\text{p}} ) - N_{\text{d}} \cdot s_{7} - T_{\text{d}} \cdot (t + r_{\text{p}} ) \\ & \quad + T_{\text{bs}} \cdot r_{\text{rt}} = 0 \\ \end{aligned}$$

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Sheet

The equilibrium of vertical forces of the sheet leads to the equation:

$$\begin{aligned} & 2 \cdot N_{\text{b1}} \cdot (\cos (\alpha /2) + \sin (\alpha /2) \cdot \mu_{3} ) + 2 \cdot N_{\text{b2}} \cdot (\cos (\alpha /2) \\ & \quad + \sin (\alpha /2) \cdot \mu_{3} ) + 2 \cdot N_{\text{r}} \cdot (\cos (\alpha /2) + \sin (\alpha /2) \cdot \mu_{3} ) \\ & \quad - 2 \cdot N_{\text{l}} \cdot (\cos (\alpha /2) - \sin (\alpha /2) \cdot \mu_{4} ) \\ & \quad - 2 \cdot N_{\text{p}} \cdot (\cos (\alpha /2) - \sin (\alpha /2) \cdot \mu_{4} ) = 0 \\ \end{aligned}$$

(34)

Appendix 4

As the static system static indefinite, one equation of the elastic deflections is used. The groove in the rotating tool (Fig. 11, magnification) concentrates the force N_r to a pre-defined position, causing an elastic deformation within the bending legs (Fig. 26).

Fig. 26

Deflection of the sheet for individual loads (Deflection is exaggerated)

The forces N_b1 and N_b2 cause an elastic deflection in the bending direction, whereas the force N_l,n operates diametrically opposite. Contact points A and B, located between the sheet and the tool, are predetermined by the position in which the lower and upper tools lock the metal sheet. These are the locations in which the elastic deflections f_A and f_B are determined. The elastic deformation depends on the attacking forces at the metal sheet, the Young’s Modulus, the geometrical moment of inertia and the lever arms, leading to Eq. 38.

$$f_{\text{A}} = g(N_{\text{b1}} ,N_{\text{b2}} ,N_{\text{l}} ,E, \, I , { }s_{\text{i}} )$$

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$$f_{\text{B}} = g(N_{\text{b1}} ,N_{\text{b2}} ,N_{\text{l}} ,E, \, I , { }s_{\text{i}} )$$

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Through the preset tool geometry, the deflections f_A and f_B are predefined in such a way that they are situated on a line ${0\bar{A}B}$ (Fig. 26). With the given lever arms, the following equation arises:

$$(s_{2} - s_{1} ) \cdot f_{\text{A}} \,\, = \,\,\,(s_{4} - s_{1} ) \cdot f_{\text{B}}$$

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4.1 Calculation of the Deflections

I = Area moment of Inertia

Deflection in Point A:

$$\begin{aligned} f_{A} & = N_{l} \cdot \left[{\frac{{(s_{4} - s_{5} - s_{1})}}{3 \cdot E \cdot I} \cdot \left({1 - 1.5 \cdot \frac{{s_{4} - s_{5} - s_{2}}}{{s_{4} - s_{5} - s_{1}}} + 0.5 \cdot \left({\frac{{s_{4} - s_{5} - s_{2}}}{{s_{4} - s_{5} - s_{1}}}} \right)^{3}} \right)} \right] \\ & \quad - N_{b1} \cdot \left[{\frac{{(s_{4} - s_{1})}}{3 \cdot E \cdot I} \cdot \left({1 - 1.5 \cdot \frac{{s_{4} - s_{2}}}{{s_{4} - s_{1}}} + 0.5 \cdot \left({\frac{{s_{4} - s_{2}}}{{s_{4} - s_{1}}}} \right)^{3}} \right)} \right] - N_{b2} \cdot \frac{{(s_{2} - s_{1})}}{3 \cdot E \cdot I} \\ \end{aligned}$$

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Deflection in Point B:

$$\begin{aligned} f_{\text{B}} & = \frac{{N_{\text{l}} \cdot (s_{4} - s_{5} - s_{1})}}{3 \cdot E \cdot I} - \frac{{N_{\text{b1}} \cdot (s_{4} - s_{1})}}{3 \cdot E \cdot I} - \frac{{N_{\text{b2}} \cdot (s_{2} - s_{1})}}{3 \cdot E \cdot I} \\ & \quad - \tan \left({\frac{{N_{\text{b2}} \cdot (s_{4} - s_{1})}}{2 \cdot E \cdot I}} \right) \cdot (s_{4} - s{}_{2}) + \tan \left({\frac{{N_{\text{l}} \cdot (s_{4} - s_{5} - s_{1})}}{2 \cdot E \cdot I}} \right) \cdot (s_{4} - s_{5}) \\ \end{aligned}$$

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