An analytical model of plate rolling force with a simple available velocity field and equal perimeter yield criterion

Abstract

To analyze plate rolling force, a simple available velocity field is first proposed in this article. Based on the velocity field, the internal deformation energy rate and the friction energy rate are analyzed by the equal perimeter yield criterion and the method of the collinear vector inner product, respectively. Finally, the rolling force accounting for the elastic flattening of the roller and the temperature rise of the rolled piece are ultimately obtained through the minimization of the total energy rate. The theoretical rolling forces are compared with actual measured data, and a good consistency is found since the maximum error between them is less than 4.7%. In addition, the discussion of different rolling conditions, such as friction factor, thickness reduction, and shape factor, upon rolling force, location of neutral angle, and stress state coefficient is also carried out.

Introduction

One significant issue in plate rolling is the accurate determination of rolling force which is the basis for the design of plate production. A three-dimensional (3D) rolling problem with a stream function velocity field has been analyzed comprehensively by Oh and Kobayashi ¹ in 1975, and the corresponding numerical solution was given. Since then, some similar numerical results have developed rapidly.^2–4 In 1980, another classical velocity field, called the weighted velocity field, was proposed by Kato et al., ⁵ and the numerical results of rolling force and torque have also been present.

The finite element method (FEM) is an important tool to analyze rolling process. Many 2D or 3D rolling processes have been analyzed using this method. A 3D rigid-plastic FEM code of shape rolling suitable for solving arbitrary intersecting surface was developed by Park and Oh. ⁶ To same computational time without losing prediction, an analysis based on the combination of FEM and slab method was carried out by Kim et al. ⁷ for 3D shape rolling. By FEM, the strip deformation behavior including the rolling forces and the plastic strain distribution was investigated by Yu et al. ⁸ The temperature distribution in the strip rolling of Mg alloy was analyzed by Ding et al. ⁹ using FEM. It is found that as the amount of thickness reduction increases, the temperature drop of surface layer first increases and then decreases. The thermo-mechanical behavior of strip rolling was investigated by Koohbor, ¹⁰ and the result shows that the thermal stress and the mechanical stress within the work roll are in a same level. The coupling between the pure mechanical model of the workpiece and the thermo-mechanical model of the rolls was developed by Cavaliere et al. ¹¹ through finite element analysis. A fast rigid-plastic FEM has been developed by Zhang et al. ¹² to analyze the strip rolling process. The predicted errors of the rolling forces are within ±10%. A predictive model of rolling force and tension profiles has been proposed by Kim et al. ¹³ In their model, the pre-deformation has been taken into account, which has improved the prediction accuracy. A finite element model of hot strip rolling has been established by Cao et al. ¹⁴ The numerical result shows that the rolling force increases as the strip width, rolling speed, rolling reduction, and friction coefficient increase. Lenard ¹⁵ carried out simulated work by FEM on cold strip rolling, and the result showed that the friction variation has apparent effect on the precision and convergence of simulations. Jiang and Tieu^16,17 reported on analysis of cold rolling of thin strips with friction variation. The effects of rolling parameters were analyzed in their study, such as work roll reductions and diameters. The cold rolling of strip was simulated by Jiang et al. ¹⁸ In the analysis, the friction change at the transverse and longitudinal directions has been considered, which is shown to affect the rolling force and strip crown. The cold rolling process has been simulated by Liu et al. ¹⁹ using 3D elastic–plastic FEM. In their simulation, the pressure between the work roll and the backup roll were taken into account.

It can be seen that many valuable results can be obtained using FEM although it is time-consuming, and little analytical investigation has been reported on rolling force and torque. As said by Abrinia and Fazlirad ²⁰ that due to the complexity of deformation equations, it is difficult to solve the strip rolling with analytical method.

In this article, a new velocity field, called the simple available velocity field, is first given. With this proposed velocity, the internal deformation energy rate, shear energy rate, and friction energy rate have been obtained using the equal perimeter (EP) yield criterion and the method of the strain vector inner product. Ultimately, the rolling force accounting for elastic flattening of the roller and the temperature rise of the rolled piece is obtained. The predicted rolling forces are validated with experimental results, and the effect of the main rolling parameters, such as friction factor, rolling reduction, and shape factor on the neutral point and stress state coefficient is also analyzed.

EP yield criterion and its specific plastic work rate

Zhang et al. ²¹ have provided the mathematical expression of the EP yield criterion and the derivation of the specific plastic work rate. The mathematical expression of the EP yield criterion in the Haigh–Westergaard stress space can be shown as

$${\sigma }_1-\frac{1}{4}{\sigma }_2-\frac{3}{4}{\sigma }_3={\sigma }_s\mathrm{if}{\sigma }_2\le \frac{1}{2}({\sigma }_1+{\sigma }_3)$$

(1)

$$\frac{3}{4}{\sigma }_1+\frac{1}{4}{\sigma }_2-{\sigma }_3={\sigma }_s\mathrm{if}{\sigma }_2\ge \frac{1}{2}({\sigma }_1+{\sigma }_3)$$

(2)

The geometric representation of the yield criterion on the π-plane is equilateral but non-equiangular dodecagon, which is shown in Figure 1.

Figure 1.

Yield loci on the π-plane.

According to Levy–Mises flow rule, the specific plastic work rate of EP yield criterion, $D({\stackrel{·}{\epsilon }}_{\mathrm{ij}})$ , can be written as ²¹

$$D({\stackrel{·}{\epsilon }}_{\mathrm{ij}})=0.5897{\sigma }_s({\stackrel{·}{\epsilon }}_{max}-{\stackrel{·}{\epsilon }}_{min})$$

(3)

It can be seen from equation (3) that the specific plastic work rate is a linear function of yield stress ${\sigma }_s$ , the maximum strain rate ${\stackrel{·}{\epsilon }}_{max}$ , and the minimum strain rate ${\stackrel{·}{\epsilon }}_{min}$ . This characteristic will provide the solving condition of the internal deformation energy rate in the next section.

Rolling energy rate

Velocity field

Due to the symmetry of rolling plate, only 1/4 deformation zone needed to be considered. The rolling plate in the deformation zone is illustrated in Figures 2 and 3, in which the coordinate axes are assigned at the center of entry cross-section. It can be seen that the initial plate with thickness 2h₀ and width 2b₀ are rolled to 2h₁ and 2b₁. The symbols h_x and b_x denote the half values of plate thickness and plate width, respectively, and the symbols h_m and b_m denote the average of h_x and b_x, respectively. The exact arc equation and its first-order derivation can be described as follows ²²

$$\begin{array}{c}{h}_x=R+h_1-\sqrt{R^2-{(l-x)}^2}\hfill \\ =h_{\alpha }=R+h_1-R\cos \alpha ;l-x=R\sin \alpha \hfill \\ h_m=\frac{R}{2}+h_1+\frac{\Delta h}{2}-\frac{R^2\theta }{2l};\hfill \\ {h\prime }_x=-\frac{l-x}{\sqrt{R^2-{(l-x)}^2}}=-\tan \alpha \hfill \end{array}$$

(4)

$$\begin{array}{c}{b}_x=b_1-\frac{\Delta b}{l^2}(l-x{)}^2\hfill \\ b_m=\frac{b_0+2b_1}{3}\hfill \end{array}$$

(5)

Figure 2.

Deforming zone for plate rolling.

Figure 3.

Half of xoy section.

A simple available velocity field of $v_x$ , $v_y$ , and $v_z$ is proposed as follows after the above assumptions, which can be shown as

$$\begin{array}{c}{v}_x=v_0(1+\lambda \frac{h_0-h_x}{h_x})\hfill \\ v_y=v_0(\lambda -1)\frac{{h\prime }_x}{h_x}y\hfill \\ v_z=v_0\frac{{h\prime }_x}{h_x}(1+\lambda \frac{h_0-h_x}{h_x})z\hfill \end{array}$$

(6)

where the value of $\lambda$ can be obtained according to the principle of mass flow

$$\lambda =1-\frac{\Delta b{h}_0}{\Delta h{b}_m}$$

Using the Cauchy equation, the strain rate components ${\stackrel{·}{\epsilon }}_x$ , ${\stackrel{·}{\epsilon }}_y$ , and ${\stackrel{·}{\epsilon }}_z$ from equation (6) can be obtained as

$$\begin{array}{l}{\stackrel{.}{\epsilon }}_x=-\lambda \frac{h_0{h^{\prime }}_x}{h_x^2}{v}_0;{\stackrel{.}{\epsilon }}_y=-(1-\lambda )\frac{{h^{\prime }}_x}{h_x}{v}_0;\\ {\stackrel{.}{\epsilon }}_z=\left[\lambda \frac{h_0{h^{\prime }}_x}{h_x^2}+(1-\lambda )\frac{{h^{\prime }}_x}{h_x}\right]v_0\end{array}$$

(7)

In equations (6) and (7), $h_x=h_0,v_x=v_0$ at x = 0; $v_y=0$ at y = 0; $v_z=0$ at z = 0; and ${\stackrel{·}{\epsilon }}_x+{\stackrel{·}{\epsilon }}_z+{\stackrel{·}{\epsilon }}_y=0$ ; therefore, the proposed velocity is a kinematically admissible one.

Internal deformation energy rate

In equation (7), taking notice of ${\stackrel{·}{\epsilon }}_{max}={\stackrel{·}{\epsilon }}_x={\stackrel{·}{\epsilon }}_1$ , ${\stackrel{·}{\epsilon }}_{min}={\stackrel{·}{\epsilon }}_z={\stackrel{·}{\epsilon }}_3$ , and substituting them into equation (3), the internal deformation energy rate can be obtained as

$$\begin{array}{cc}\multicolumn{1}{c}{{\stackrel{·}{W}}_i}& ={\int }_{V}D\left({\stackrel{·}{\epsilon }}_{\mathrm{ij}}\right)\mathrm{dV}\hfill \\ \multicolumn{1}{c}{}& =4{\int }_0^l{\int }_0^{b_x}{\int }_0^{h_x}0.5897{\sigma }_s({\stackrel{·}{\epsilon }}_{max}-{\stackrel{·}{\epsilon }}_{min})\mathrm{dxdydz}\hfill \\ \multicolumn{1}{c}{}& =\frac{2.3588{\sigma }_{s}U}{h_0}((1-\lambda )(h_0-h_1)+2\lambda h_0\ln \frac{h_0}{h_1})\hfill \end{array}$$

(8)

where $U=v_0{h}_0{b}_0=v_x{h}_x{b}_x=v_n{h}_n{b}_n=v_1{h}_1{b}_1$

It has to be emphasized here that the linear specific plastic work rate of the EP yield criterion has the vital role in obtaining the analytical internal deformation energy rate.

Friction energy rate

In Figure 4, it can be seen that the tangential velocity discontinuity in interface could be expressed as

$$\begin{array}{cc}\multicolumn{1}{c}{\left|\Delta v_f\right|}& =\sqrt{\Delta v_x^2+\Delta v_y^2+\Delta v_z^2}\hfill \\ \multicolumn{1}{c}{}& =\sqrt{v_y^2+{(v_R\cos \alpha -v_x)}^2+{(v_R\sin \alpha -v_x\tan \alpha )}^2}\hfill \\ \multicolumn{1}{c}{}& =\sqrt{v_y^2+{(v_R-v_x\sec \alpha )}^2}\hfill \\ \multicolumn{1}{c}{\Delta v_f}& =\Delta v_{x}i+\Delta v_{y}j+\Delta v_{z}k=(v_R\cos \alpha -v_x)i\hfill \\ \multicolumn{1}{c}{}& +v_{y}j+(v_R\sin \alpha -v_x\tan \alpha )k\hfill \end{array}$$

(9)

where the friction shear stress along the tangential direction of the interface, ${\tau }_f=\mathrm{mk}$ , is the co-line vector relative with the tangential velocity discontinuity $\Delta v_f$ . The co-line vector integral for the friction energy rate can be expressed as

$$\begin{array}{cc}\multicolumn{1}{c}{{\stackrel{·}{W}}_f}& =4{\int }_0^l{\int }_0^{b_x}{\tau }_f\left|\Delta v_f\right|\mathrm{dF}=4{\int }_0^l{\int }_0^{b_x}{\tau }_f·\Delta v_f\mathrm{dF}\hfill \\ \multicolumn{1}{c}{}& =4{\int }_0^l{\int }_0^{b_x}({\tau }_{\mathrm{fx}}\Delta v_x+{\tau }_{\mathrm{fy}}\Delta v_y+{\tau }_{\mathrm{fz}}\Delta v_z)\mathrm{dF}\hfill \\ \multicolumn{1}{c}{}& =4\mathrm{mk}{\int }_0^l{\int }_0^{b_x}(\Delta v_x\cos \alpha +\Delta v_y\cos \beta +\Delta v_z\cos \gamma )\mathrm{dF}\hfill \end{array}$$

(10)

where the black dot refers to the operation of dot product; $\alpha$ , $\beta$ , and $\gamma$ denote the angles between the $\Delta v_f$ (or ${\tau }_f$ ) and the x, y, and z axes.

Figure 4.

Co-line vectors at interface.

Since the direction of the tangential velocity discontinuity coincides with that of roller surface, the direction cosines along the tangential direction can be calculated as

$$\begin{array}{l}\cos \alpha =\pm \frac{\sqrt{R^2-{(l-x)}^2}}{R},\cos \beta =0,\\ \cos \gamma =\pm \frac{l-x}{R}=\pm \sin \alpha \end{array}$$

(11)

Following the Pavlov projection principle, the contact surface and its tangential velocity discontinuity can be written as

$$\begin{array}{cc}\multicolumn{1}{c}{{\stackrel{·}{W}}_f}& =4{\int }_0^l{\int }_0^{b_x}{\tau }_f\left|\Delta v_f\right|\mathrm{dF}=4{\int }_0^l{\int }_0^{b_x}{\tau }_f·\Delta v_f\mathrm{dF}\hfill \\ \multicolumn{1}{c}{}& =4{\int }_0^l{\int }_0^{b_x}({\tau }_{\mathrm{fx}}\Delta v_x+{\tau }_{\mathrm{fy}}\Delta v_y+{\tau }_{\mathrm{fz}}\Delta v_z)\mathrm{dF}\hfill \\ \multicolumn{1}{c}{}& =4\mathrm{mk}{\int }_0^l{\int }_0^{b_x}\left|\Delta v_f\right|\sec \alpha \mathrm{dxdy}\hfill \end{array}$$

(12)

On the roll surface, taking notice of $\mathrm{dz}/\mathrm{dy}=0$ and substituting equations (11) and (12) into equation (10) and integrating, it yields

$$\begin{array}{cc}\multicolumn{1}{c}{{\stackrel{·}{W}}_f}& =4{\int }_0^l{\int }_0^{b_x}{\tau }_f\Delta v_f\mathrm{dF}=4\mathrm{mk}{\int }_0^l{\int }_0^{b_m}|v_R-v_x\sec \alpha |\sec \alpha \mathrm{dxdy}=4\mathrm{mk}{\int }_0^l{b}_m|v_R-v_x\sec \alpha |\sec \alpha \mathrm{dx}\hfill \\ \multicolumn{1}{c}{}& =4\mathrm{mkR}[v_R{b}_m(-2{\alpha }_n+\theta )+\frac{U{b}_m}{h_0{b}_0}{\mu }_b\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+\frac{U{b}_m}{h_0{b}_0}{\mu }_f\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})]\hfill \end{array}$$

(13)

where ${\mu }_f=(1+\lambda \left(\frac{h_0-h_{\mathrm{mf}}}{h_{\mathrm{mf}}}\right)),{\mu }_b=(1+\lambda \left(\frac{h_0-h_{\mathrm{mb}}}{h_{\mathrm{mb}}}\right)),h_{\mathrm{mf}}\stackrel{·}{=}\frac{h_{{\alpha }_n}+2h_1}{3},h_{\mathrm{mb}}\stackrel{·}{=}\frac{h_0+2h_{{\alpha }_n}}{2}$ , and $k={\sigma }_s/\sqrt{3}$ is the yield shear stress.

Shear energy rate

In the exit section of the deformation zone, the velocity field could be expressed as

$$x=l,h{\prime }_x=0,{v_z|}_{x=l}=\Delta v_z{|}_{x=l}={\begin{array}{c}{v}_y\hfill \end{array}|}_{x=l}=\Delta v_y{|}_{x=l}=0$$

Thus, it indicates that along the exit section, no shear power is dissipated. Here, it is considered that if the double barreling in slab rough rolling occurs, then a parabolic distribution of $v_z$ along z-direction can be assumed. Thus, on the entry section, the following average velocity components and the shear energy rate can be obtained as

$${\overline{v}}_z=\frac{1}{h_0}{\int }_0^{h_0}{v}_z\mathrm{dz}=-\frac{\tan \theta v_0}{3}$$

(14)

$$\begin{array}{l}{\overline{v}}_y=\frac{1}{b_0}{\displaystyle {\int }_0^{b_0}{v_y|}_{x=0}}dy=\frac{1}{b_0}{\displaystyle {\int }_0^{b_0}{v}_0\left(\lambda -1\right)\frac{-\tan \theta }{h_0}y}dy\\ =\frac{1}{3}{v}_0{b}_0\left(\lambda -1\right)\frac{-\tan \theta }{h_0}\end{array}$$

(15)

$$\begin{array}{l}{\stackrel{.}{W}}_{s0}=4k{\displaystyle {\int }_0^{b_0}{\displaystyle {\int }_0^{h_0}\left({\sqrt{\Delta {\overline{v}}_y^2+\Delta {{\overline{v}}_z}_z^2}|}_{x=0}\right)dzdy}}\\ =\frac{4}{3}kU\tan \theta \sqrt{1+{\left(1-\lambda \right)}^2{\left(\frac{h_0}{b_0}\right)}^2}\end{array}$$

That is to say, the total shear energy dissipation rate of entry section can be expressed as

$${\stackrel{·}{W}}_s={\stackrel{·}{W}}_{s0}=\frac{4}{3}\mathrm{kU}\tan \theta \sqrt{1+{(1-\lambda )}^2{\left(\frac{b_0}{h_0}\right)}^2}$$

(16)

Total energy rate

Substituting equations (8), (13), and (16) into $\mathrm{\Phi }={\stackrel{·}{W}}_i+{\stackrel{·}{W}}_f+{\stackrel{·}{W}}_s$ yields the analytical solution of the total energy rate as

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{\Phi }=\frac{2.3588{\sigma }_{s}U{b}_m}{h_0{b}_0}(2\lambda h_0\ln \frac{h_0}{h_1}+(1-\lambda )(h_0-h_1))+2\mathrm{kU}\tan \theta \sqrt{1+{(1-\lambda )}^2{\left(\frac{b_0}{h_0}\right)}^2}}\\ \multicolumn{1}{c}{+4\mathrm{mkR}[v_R{b}_m(-2{\alpha }_n+\theta )+\frac{U{b}_m}{h_0{b}_0}{\mu }_b\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+\frac{U{b}_m}{h_0{b}_0}{\mu }_f\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})]}\end{array}$$

(17)

Defining $\epsilon =\ln (h_0/h_1)$ and differentiating the total energy $\mathrm{\Phi }$ with respect to neutral angle ${\alpha }_{\mathrm{n}}$ and setting it be zero, that is, $\partial \mathrm{\Phi }/\partial {\alpha }_n=0$ , then the following equation can be obtained as

$$\frac{d\mathrm{\Phi }}{d{\alpha }_n}=\frac{\partial {\stackrel{·}{W}}_i}{\partial {\alpha }_n}+\frac{\partial {\stackrel{·}{W}}_f}{\partial {\alpha }_n}+\frac{\partial {\stackrel{·}{W}}_s}{\partial {\alpha }_n}=0$$

(18)

From equations (8), (13), and (15), the derivation can be obtained as follows

$$\begin{array}{c}\frac{\partial {\stackrel{·}{W}}_i}{\partial {\alpha }_n}=\frac{2.3588{\sigma }_{s}N{b}_m}{h_0{b}_0}(2\lambda h_0\ln \frac{h_0}{h_1}+(1-\lambda )(h_0-h_1))\hfill \\ \frac{\partial {\stackrel{·}{W}}_f}{\partial {\alpha }_n}=4\mathrm{mkR}\left[\begin{array}{c}-2v_R{b}_m+\frac{N{b}_m}{h_0{b}_0}({\mu }_b\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+{\mu }_f\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2}))\hfill \\ \multicolumn{1}{c}{-\frac{\lambda R\sin {\alpha }_{n}U{b}_m}{2b_0}(\frac{1}{{h_{\mathrm{mb}}}^2}\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+\frac{1}{{h_{\mathrm{mf}}}^2}\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2}))+\frac{U{b}_m}{h_0{b}_0}\frac{{\mu }_b+{\mu }_f}{\cos {\alpha }_n}}\end{array}\right]\hfill \\ \frac{\partial {\stackrel{·}{W}}_s}{\partial {\alpha }_n}=2k\tan \theta \frac{N}{h_0{b}_0}\sqrt{1+{(1-\lambda )}^2{\left(\frac{b_0}{h_0}\right)}^2}\hfill \end{array}$$

(19)

where $\partial U/\partial {\alpha }_n=v_R{b}_{m}R\sin 2{\alpha }_n-v_R{b}_m(R+h_1)\sin {\alpha }_n=N.$

Solving equation (17) yields

$$\begin{array}{cc}\multicolumn{1}{c}{m=}& (\frac{2.3588{\sigma }_{s}N{b}_m}{h_0{b}_0}(2\lambda h_0\ln \frac{h_0}{h_1}+(1-\lambda )(h_0-h_1))+2k\tan \theta \frac{N}{h_0{b}_0}\sqrt{1+{(1-\lambda )}^2{\left(\frac{b_0}{h_0}\right)}^2})/4\mathrm{kR}\hfill \\ \multicolumn{1}{c}{}& {\left\{\begin{array}{c}-2v_R{b}_m+\frac{N{b}_m}{h_0{b}_0}({\mu }_b\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+{\mu }_f\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2}))+\hfill \\ \multicolumn{1}{c}{\frac{U{b}_m}{h_0{b}_0}[\frac{-\lambda h_{0}R\sin {\alpha }_n}{2{h_{\mathrm{mb}}}^2}\ln \frac{\tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+\frac{{\mu }_b+{\mu }_f}{\cos {\alpha }_n}+\frac{-\lambda h_{0}R\sin {\alpha }_n}{2{h_{\mathrm{mf}}}^2}\ln \tan (\frac{\pi }{4}+\frac{{\alpha }_n}{2})]}\end{array}\right\}}^{-1}\hfill \end{array}$$

(20)

Using the minimum total energy rate, the corresponding rolling torque $M$ , rolling force $F$ , and stress effective factor $n_{\sigma }$ can be obtained as

$$M=\frac{R}{2{\nu }_R}{\mathrm{\Phi }}_{min},F=\frac{M}{\chi \sqrt{2R\Delta h}},n_{\sigma }=\frac{\overline{p}}{2k}=\frac{F}{4b_m\mathrm{lk}}$$

(21)

where the arm factor $\chi$ can refer to the following equation ²²

$$\chi =0.39+0.295·e^{-0.193·R/(2h_m)}$$

(22)

Validation and discussions

Rolling experimental data of rolling force from one domestic factory have been adopted here to check the correctness of the derived equation (21). The dimension of the rolling billet in the rolling is 320 × 2050 × 3250 (mm³), and the diameter of the roller is 1070 mm. In the first rolling pass, the billet is rolled to 299 mm, and then is turned 90° for the remaining broadside rolling passes. The deformation resistance of the Q345 steel determined by Chen et al. ²³ takes the following form

$$\begin{array}{c}{\sigma }_s=6310.7·e^{(-2.62T/1000-0.669e)}{\stackrel{·}{\epsilon }}^{0.115}{\epsilon }^{0.407}\hfill \\ T=t+237\hfill \end{array}$$

(23)

Equation (23) indicates that the flow stress is a function of applied true strain $\epsilon$ , strain rate $\stackrel{·}{\epsilon }$ , and rolling temperature T.

In fact, the rolling temperature in the above-mentioned passes can be obtained according to the temperature rise and fall models. The temperature rise model during the deformation zone can be expressed as ²⁴

$$\Delta T_h=\frac{\eta \int \sigma d\epsilon }{\rho C}=\frac{\eta {\sigma }_s\epsilon }{\rho C}$$

(24)

and the temperature fall model in air cooling can be described as ²⁵

$$\Delta T_c=\frac{2{\sigma }_s\epsilon }{\rho C{h}_m}{\left(\frac{T_{\mathrm{IN}}}{100}\right)}^4\Delta \tau$$

(25)

In this article, the influence of roll flattening is taken into account, and the roller radius can be described as ²⁵

$$R=R_0(1+2.2\times {10}^{-5}\frac{F}{4b_m\Delta h})$$

(26)

where $R_0$ is the initial roller radius.

Due to the coupling relationship between the roll flattening and rolling force, the calculation procedure of roll force is iterative. The computing flow chart is shown in Figure 5, and the condition of convergence is $\frac{|R_i-R_{i-1}|}{R_i}\le 0.001$ . The actual rolling force is obtained by a pressure sensor equipped on the hydraulic cylinder under which the pressure is depressed. The comparison results are shown in Table 1 and Figure 6.

Figure 5.

Flow chart of the calculation.

Table 1.

Comparison of rolling separating forces calculated by equation (21) with the measured ones.

Pass no.	vR m/s	$h_0$ (mm)	$\epsilon$ (%)	$T$ (°C)	$\Delta \tau$ (s)	$R$ (%)	Measured force (kN)	Calculated force (kN)	Error △ (%)
2	–2.48	312.60	6.6	980.3	0.38	0.5561	43,712	44,338	1.4
3	2.46	290.40	7.1	986.5	0.48	0.5562	43,790	44,748	2.2
4	–2.50	268.32	7.6	990.2	0.64	0.5564	44,885	45,385	1.1
5	2.50	245.98	8.3	996.8	0.83	0.5565	45,812	46,251	1.0
6	–2.50	223.54	9.1	1003.6	1.14	0.5559	45,479	47,634	4.7

Figure 6.

Comparison of the present rolling force with measured ones.

As illustrated in Table 1, the predicted rolling forces coincide well with the experimental values since the maximum error between them is less than 4.7%. However, it can be observed that all the predicted values are bigger than the experimental ones, which is reasonable since the present results are based on the upper bound method.

Figure 7 shows that ${\stackrel{·}{W}}_i$ and ${\stackrel{·}{W}}_s$ are the main portions of the ${\mathrm{\Phi }}_{min}$ , and the ${\stackrel{·}{W}}_f$ is very small, which indicates that the friction energy is not the key factor affecting the total energy rate for hot plate rolling.

Figure 7.

Ratio of ${\stackrel{·}{W}}_i$ , ${\stackrel{·}{W}}_s$ , and ${\stackrel{·}{W}}_f$ to ${\mathrm{\Phi }}_{min}$ with different reduction r.

In Figure 8, the location of neutral points $x_n/l$ is a function of friction factor m and reduction $r$ . When m decreases or $r$ increases, the neutral point moves toward the exit plane. Moreover, the neutral point changes obviously with a small change in friction $m<0.3$ , which indicates that the rolling process may become unstable in this range of frictions.

Figure 8.

Effect of m and r on the location of neutral points.

Figure 9 illustrates the effect of shape factor l/(2h_m) on the stress state coefficient $n_{\sigma }$ . It can be seen that for a specific m, the stress state coefficient increased with the increasing l/(2h_m), while m has insignificant effect on the stress state coefficient, which shows the same rule as shown in Figure 7.

Figure 9.

Effect of $l/2h_m$ and m on $n_{\sigma }$ .

Figure 10 illustrates that the deformation factors of r and $R/h_0$ have significant influences on the stress coefficient $n_{\sigma }$ . It can be seen that $n_{\sigma }$ increased with increasing r and $R/h_0$ .

Figure 10.

Effect of r and $R/h_0$ on $n_{\sigma }$

Conclusion

1. In order to describe the plate rolling, a simple available velocity field which satisfies kinematical admissible condition is proposed. Through this velocity and by using the EP yield criterion and the method of the strain vector inner product, the analytical solution of rolling force is ultimately obtained.

2. A low value of l/h produces a high value of stress state coefficient, and m has insignificant effect on the stress state coefficient since the friction energy rate has the least contribution to the total energy function.

3. With the decreasing m or increasing r, the neutral point moves to the exit plane. If the value of m is less than 0.3, unstable rolling process will occur.

4. Due to the consideration of the elastic flattening of the rollers and the temperature rise of rolled piece, the present calculated rolling forces coincide well with measured ones, and all the relative errors are less than 4.7%.