# Optimization Design of the Lower Rocker Arm of a Vertical Roller Mill Based on ANSYS Workbench
## 2 Finite Element Analysis
### Load Calculation  
The force system needs to be analyzed before calculating the force on the lower rocker arm. In Figure 2a, the marked components 1, 2, 3 and 4 in the figure are represented as the lower rocker arm, upper rocker arm, grinding roller and rocker arm shaft, respectively. The rocker shaft is connected to the bearing base through the bearing, the upper rocker arm and lower rocker arm are connected to the rocker arm shaft through the expansion sleeves, the tail of the lower rocker arm is connected to the hydraulic device, and the upper rocker arm is rigidly connected to the grinding roller. Assuming that the VRM works under ideal conditions, the force acting on the force system can be appropriately simplified. As shown in Figure 2b, \(F_{\text{H}}\) is the pulling force provided by the hydraulic cylinder, and \(F_{\text{ax}}\), \(F_{\text{ay}}\) and \(F_{\text{ax}}\) are loads in three directions on the connection of the two rocker-arm bearings respectively. \(F_{\text{ax1}}\) and \(F_{\text{ax2}}\) represent the force component of \(F_{\text{ax}}\) acting on the joint of the two rocker-arm bearings. \(F_{\text{bx}}\) is the sliding friction between the grinding roller and the material, \(F_{\text{by}}\) is the rolling friction in the tangential direction of the outer circle of the grinding roller, while \(F_{\text{bz}}\) is the reaction force of the grinding disc to the grinding roller (including the self-weight of the grinding roll G = 1.30 \(\times\) 10\({}^{5}\) N). Figure 1: Grid division of the lower rocker arm. Input the known parameters into Formula (1) to obtain the pulling force \(F_{\rm H}\) of the hydraulic cylinder on the rocker arm as follows:  
\[F_{\rm H}=P\times A=1.308\times 10^{6}N \tag{1}\]  
where \(A\) is the rod cavity area of the hydraulic cylinder, taken as 0.109 m\({}^{2}\), and \(P\) is the working pressure of the hydraulic cylinder, taken as 12 MPa [10]. Input the known parameters into Formula (2) to obtain the sliding friction force \(F_{\rm bx}\) between the grinding roller and the material as follows:  
\[F_{\rm bx}=f\times F_{\rm bz}=0.25F_{\rm bz} \tag{2}\]  
where \(f\) is the sliding friction coefficient between the grinding roller and the material, which is taken as 0.25 [11]. Input the known parameters into Equation (3); the reaction force \(F_{\rm bz}\) of grinding disc to grinding roll is obtained as follows:  
\[F_{\rm by}=f_{\rm r}\times F_{\rm bz}=0.05F_{\rm bz} \tag{3}\]  
where \(f_{\rm r}\) is the rolling friction coefficient between the grinding roller and the material, which is taken as 0.05 [11]. According to the equilibrium conditions:  
\[\left\{ \begin{array}{l}F_{\rm ax}-F_{\rm bx}-F_{\rm H}\times\cos\theta=0\\ F_{\rm ay}-F_{\rm by}=0\\ F_{\rm az}+F_{\rm bz}-F_{\rm H}\times\sin\theta=0\\ F_{\rm H}\times L_{\rm H}=F_{\rm B}\times L_{\rm B}\end{array} \right. \tag{4}\]  
In Figure 3, \(L_{1}\) = 1536 mm, \(L_{2}\) = 1287 mm, \(L_{3}\) = 54.3 mm, \(L_{4}\) = 1268 mm, \(\theta\) = 58\({}^{\circ}\), and the weight of the grinding roller G = 1.30 \(\times\) 10\({}^{5}\) N.  
Input the parameters into Equation (4) to obtain: \(F_{\rm ax}\) = 998 KN, \(F_{\rm ay}\) = 60.8 KN, \(F_{\rm az}\) = -129 KN, \(F_{\rm ax1}\) = 572.7 KN, \(F_{\rm ax2}\) = 425.3 KN, \(F_{\rm bx}\) = 304 KN, \(F_{\rm by}\) = 60.8 KN, and \(F_{\rm bz}\) = 1216 KN. As shown in Figure 3, \(F_{\rm cx}\), \(F_{\rm cy}\), and \(F_{\rm cz}\) are the forces exerted by the upper and lower rocker arms at the connection holes; \(F_{\rm cx1}\) and \(F_{\rm cx2}\) represent the force component distributed by \(F_{\rm cx}\) to the two connecting holes. Figure 2: Force analysis of the force system of the VRM: (**a**) 3D structure of the force application system; (**b**) force diagram of force application system. According to the equilibrium conditions:  
\[\left\{ \begin{array}{l}F_{\rm cy}-F_{\rm by}=0\\ F_{\rm cz}-F_{\rm bz}=0\\ F_{\rm cx1}+F_{\rm cz2}=F_{\rm cx}\\ \frac{(F_{\rm cx1}-F_{\rm cz2})}{2}\times L_{4}=F_{\rm by}\times L_{1}\end{array} \right.