# System Design and Experimental Study of a Four-Roll Bending Machine
## 3 Design
### Force Analysis of the Pre-Bending Process  
During the pre-bending operation of a four-roll bending machine, the two side rollers exhibit an asymmetric configuration, causing the line of action of force to rotate at a specific angle relative to the rolling operation. The geometric configuration under pre-bending conditions is illustrated in Figure 8.  
When the forming radius decreases, the force acting on each roller increases. To accommodate the maximum design load, 45 steel is selected as the target material. According to the design specifications, when the forming radius is 1000 mm, the pre-springback neutral layer radius \(R_{T}\) can be calculated as 799 mm using Equation (1).  
\[R_{f}=\frac{R_{T}}{1-3\frac{R_{T}\sigma_{y}}{TE}+4\frac{R_{T}^{3}\sigma_{y}^{3} }{T^{3}E^{3}}}. \tag{1}\]  
where, \(R_{T}\) denotes the radius before springback, \(R_{f}\) denotes the radius after springback, \(T\) denotes the profile thickness, \(E\) denotes the elastic modulus, and \(\sigma_{y}\) denotes the yield strength. Section Modulus of the Profile:  
\[W=\frac{HT}{6}=1333.3\,(\mathrm{mm^{3}}). \tag{2}\]  
where, \(H\) denotes the square of the profile height, and the profile height is 20 mm. Maximum Bending Moment for Deformation:  
\[M=(K_{1}+\frac{K_{0}T}{2R_{T}})W\sigma_{S}=8.14\times 10^{5}\,(\mathrm{N\cdot mm }). \tag{3}\]  
Figure 8: Pre-bending Force Model.  
where, the relative strengthening coefficient \(K_{0}\) is taken as 17.6, the cross-sectional shape coefficient \(K_{1}\) is taken as 1.5, the thickness \(T\) of the rolled section is 20 mm, and the yield strength \(\sigma_{5}\) of 45 steel is 355 MPa. Length of the Residual Straight Edge [19]:  
\[C^{\prime}=2T=40\,(\text{mm}). \tag{4}\]  
Therefore,  
\[O_{2}C=\left[1+\frac{D_{2}}{(2R_{T}+T)}\right]C^{\prime}=42.72\,(\text{mm}). \tag{5}\]  
where, the diameter \(D_{2}\) of the lower roller is 110 mm. In triangle \(OO_{2}C\):  
\[OO_{2}=R_{T}+\frac{T}{2}+\frac{D_{2}}{2}. \tag{6}\]  
\[\gamma=\arcsin(\frac{O_{2}C}{OO_{2}})=2.83^{\circ}. \tag{7}\]  
\[OC=\sqrt{\left(OO_{2}\right)^{2}-\left(O_{2}C\right)^{2}}=\sqrt{\left(R_{T}+ \frac{T}{2}+\frac{D_{2}}{2}\right)^{2}-\left(O_{2}C\right)^{2}}. \tag{8}\]  
Therefore,  
\[O_{1}C=OC-OO_{1}=OC-(R_{T}-\frac{T}{2}-\frac{D_{1}}{2}). \tag{9}\]  
where, the diameter \(D_{1}\) of the upper roller is 110 mm. In triangle \(O_{1}O_{2}C\):  
\[\phi=\arctan(\frac{O_{2}C}{O_{1}C})=18.33^{\circ}. \tag{10}\]  
In triangle \(NOD\):  
\[\sin\angle GOD=\sin(\theta-\phi)=\frac{OD}{GO}. \tag{11}\]  
where, the tilt angle \(\theta\) of the side roller is \(25^{\circ}\). In triangle \(O_{3}OD\):  
\[\sin\alpha=\frac{OD}{OO_{3}}. \tag{12}\]  
Therefore,  
\[\frac{\sin\alpha}{\sin(\theta-\phi)}=\frac{GO}{OO_{3}}=\frac{L_{1}+\frac{D_{ 2}}{2}+\frac{T}{2}-R_{T}}{R_{T}+\frac{T}{2}+\frac{D_{3}}{2}}. \tag{13}\]  
In triangle \(GO_{1}F\):  
\[\sin(\theta-\phi)=\frac{O_{1}F}{L_{1}}. \tag{14}\]  
In triangle \(AO_{1}F\):  
\[\sin\theta=\frac{O_{1}F}{L}. \tag{15}\]  
where, \(L\) is 323.81 mm. Therefore,  
\[\frac{\sin\theta}{\sin(\theta-\phi)}=\frac{L_{1}}{L}. \tag{16}\]  
Simultaneous Equations (13) and (16):  
\[\alpha=3.42^{\circ}. \tag{17}\]  
\[\beta=\theta+\alpha-\phi=10.09^{\circ}. \tag{18}\]Force on the Upper Roller (\(F_{1}\)):  
\[F_{1}=\frac{M}{R_{T}+\frac{T}{2}}(\frac{1}{\tan\gamma}+\frac{1}{\tan\beta})=26008.8 \,(\mathrm{N}). \tag{19}\]  
Force on the Lower Roller (\(F_{2}\)):  
\[F_{2}=\frac{M}{(R_{T}+\frac{T}{2})\sin\gamma}=20379\,(\mathrm{N}). \tag{20}\]  
Force on the Side Roller (\(F_{3}\)):  
\[F_{3}=\frac{M}{(R_{T}+\frac{T}{2})\sin\beta}=5743\,(\mathrm{N}). \tag{21}\]