Modeling of rolling force of ultra-heavy plate accounting for gradient temperature

Abstract

In this paper, the nonlinear specific plastic power of the Mises criterion is integrated analytically to establish the rolling force model of gradient temperature rolling for an ultra-heavy plate by a new method called the root vector decomposition method. Firstly, the sinusoidal velocity field is proposed in terms of the characteristics of metal flow during ultra-heavy plate rolling, which satisfies the kinematically admissible condition. Meanwhile, the characteristics of the temperature distribution along the thickness direction of the plate during the gradient temperature rolling is described mathematically. Based on the velocity field and the temperature distribution expression, the rolling energy functional is obtained by using the root vector decomposition method, and the analytical solution of rolling force is derived according to the variational principle. Through comparison and verification, the rolling force model solved by the root vector decomposition method in this paper is in good agreement with the measured one, and the maximum error of the rolling force is just 10.21%.

Introduction

In recent years, gradient temperature rolling, as an advanced rolling technology which can effectively improve the quality of ultra-heavy plate, has received extensive attention. It is different from traditional uniform temperature rolling since rapid cooling is introduced. Therefore, the temperature of the plate is always shown to be low on surface and high in core. The research on gradient temperature rolling technology can be traced back to the Super-OLAC system developed by JFE Company in 1998. ¹ By using this technology, the product quality has been significantly improved, and the yield of finished products has been increased by about 20%. ² Many enterprises and scholars in China also have a strong interest in it, such as Tangshan Wenfeng Shanchuan Wheel Co. Ltd, ³ Xingcheng Special Steel, ⁴ Wuhan Iron and Steel Company, ⁵ and many other companies. In order to meet the needs of production, many researchers have carried out various researches. Through experiments, Zhao ⁶ found that the temperature gradient in the thickness direction promoted the core deformation of the thick plate. Shou-yuan et al. ⁷ analyzed the temperature and strain distribution in a thick plate during gradient temperature rolling. Their results show that the uniform deformation in the thickness direction is promoted due to the temperature gradient. Wei et al.^8,9 found that the strain at the core of the plate and the cumulative deformation of the plate are increased during gradient temperature rolling. In the aspect of mechanical properties, the ultra-heavy plate with gradient temperature rolling has better toughness than that with uniform temperature rolling, and the mechanical properties along the thickness direction are improved. The above research and application have proved that gradient temperature rolling can effectively improve the production quality. However, due to the lack of corresponding mathematical model, it is difficult to guarantee the product quality in actual production. ¹⁰ Therefore, it is of great significance to establish the rolling force model for the application of gradient temperature rolling.

Sellars ¹¹ found that temperature is an important variable in the process of metal deformation. Due to the uneven temperature distribution in the thickness direction of the plate during the gradient temperature rolling, the rolling force will vary compared with the uniform temperature rolling. He ¹² compared the rolling force of a thick plate with gradient temperature and that with uniform temperature through experiments. It is found that the rolling force is always greater than that of the uniform rolling because of the increase of the average deformation resistance of the plate. Therefore, it is necessary to establish a new rolling force model according to the characteristics of gradient temperature rolling. Phaniraj et al. ¹³ simulated the hot plate rolling process with gradient temperature to obtain the temperature field and the rolling pressure for different stands. Aiyedun et al. ¹⁴ established a finite difference model to simulate the temperature distribution in the gradient temperature rolling. By analyzing the temperature distribution data and the measured rolling force and torque, they found that with the decrease of strain rate, the influence of temperature gradient on the rolling force and torque is gradually intensified. Zhang et al. ¹⁵ established a 3D rigid-plastic thermo-mechanical finite element model to study the edge deformation and metal flow of a heavy plate in the rolling process, and obtained rolling force data with large temperature gradient, which fits well with the measured data. Although the relevant rolling force data can be obtained through simulation in the above research, the simulation process was time-consuming and difficult to be applied in actual production. In order to speed up the calculation in gradient temperature rolling process and guarantee the accuracy of temperature distribution in thickness direction, Ding et al. ¹⁶ investigated the temperature control technology by finite difference scheme with thickness unequally partitioned method. However, this method is not universal, and different models need to be established under different conditions. Its accuracy fluctuates greatly due to the limitation of the level of operator, so it is difficult to be applied on-line. In order to obtain a rolling force model that is generally applicable to the gradient temperature rolling process, Li ¹⁷ modified the traditional rolling force model through the data by finite element simulation and obtained a simple formula to calculate the gradient temperature rolling force. However, due to the absence of consideration of deformation permeability effect, there is a fiercely fluctuate of the predicted results. Wang et al. ¹⁸ obtained the gradient temperature rolling force model of an ultra-heavy plate based on the engineering method. However, due to the use of many assumptions, the errors of the model are big, and it is difficult to meet the requirements of the precise control of rolling process. Jiang et al. ¹⁹ used the engineering method to divide the rolling of snake/gradient temperature rolling into three layers for calculation. However, this method cannot reflect the effect of temperature distribution on heavy plate during gradient temperature rolling process. At present, the analytical method of gradient temperature rolling force is mostly adopted by engineering method. A large number of assumptions are made in the derivation process, and the results need to be further improved.

In order to solve the above problems, an analytical solution of gradient temperature rolling force is established by using energy method, since this method can provide an upper bound solution with good calculation accuracy in terms of a reasonable velocity field, and can disclose the relationship between the rolling force and process parameters well. However, due to the difficulty of solving the nonlinear specific plastic work rate, few literatures using this method are reported. In order to solve this problem, a new method, called the Root vector decomposition method, is firstly proposed in the present paper. Meanwhile, the sinusoidal velocity field is established according to the characteristics of the metal flow during the rolling process, the temperature distribution expression is established according to the characteristics of the temperature distribution of the ultra-heavy plate. Ultimately, the analytical solution of gradient temperature rolling force is obtained, and the prediction accuracy of the model is verified by comparing with the experimental data.

Rolling force model

Sinusoidal velocity field

When the ratio of plate width to thickness satisfies $b/h\ge 10$ , the metal flow in the width direction can be ignored and the rolling process can be regarded as a two-dimensional plane deformation problem.

The rolling deformation diagram of an ultra-heavy plate is shown in Figure 1. Taking the center of the plate entrance section as the origin of coordinates, the length direction of the plate is along the x axis and the thickness direction is along the z axis. The point O is in the rolled center, R is the roll radius, $\theta$ is the contact angle, l is the projected length of the deformation zone, $h_0$ and $h_1$ are respectively the inlet and outlet thickness of the rolled piece, $v_0$ and $v_1$ are respectively the inlet and outlet velocity of the rolled piece. Since the deformation zone is symmetrically distributed, only one quarter of the deformation zone is taken for research.

Figure 1.

Rolling deformation diagram.

Denote the plate thickness at the point x away from the entrance section by $h_x$ , then the contact arc equation is

$$h_x=R+h_1-\sqrt{R^2-{(l-x)}^2}$$

(1)

At this point denote the circle angle corresponding to arc BE by $\alpha$ , then the contact arc parametric equation is

$$\begin{array}{c}{h}_x=R+h_1-R\cos \alpha \hfill \\ l-x=-R\sin \alpha ,\mathrm{d}x=-R\cos \alpha \mathrm{d}\alpha \hfill \end{array}$$

(2)

The first and second derivatives parametric equations of $h_x$ are

$$\begin{array}{c}h{\prime }_x=\frac{\mathrm{d}{h}_x}{\mathrm{d}x}=\frac{l-x}{\sqrt{R^2-{(l-x)}^2}}=-\tan \alpha \hfill \\ h\prime {\prime }_x=\frac{\mathrm{d}h{\prime }_x}{\mathrm{d}x}={\left(R{\cos }^3\alpha \right)}^{-1}\hfill \end{array}$$

(3)

As shown in Figure 1, the boundary condition is

$$\{\begin{array}{c}x=0,\alpha =\theta ,h_x=h_0,h{\prime }_x=-\tan \theta ,v_x=v_0\hfill \\ \multicolumn{1}{c}{x=l,\alpha =0,h_x=h_1,h{\prime }_x=0,v_x=v_1,v{\prime }_x=0}\end{array}$$

(4)

In the rolling process, since $b/h\ge 10$ , the change of plate width is ignored in the rolling process, so $y\equiv b$ .

It can be found that the horizontal velocity of the rolling work piece increases continuously from the inlet to the outlet and the slope of the horizontal component of the velocity at the outlet is zero. As a result, the following two-dimensional velocity field can be obtained by assuming that the metal flow velocity along the x-axis in the rolling process follows the sinusoidal function distribution

$$\{\begin{array}{c}{v}_x=v_1\sin [(\frac{\pi }{2}-\arcsin \frac{v_0}{v_1})\frac{x}{l}+\arcsin \frac{v_0}{v_1}]\hfill \\ \multicolumn{1}{c}{v_y=0}\\ \multicolumn{1}{c}{v_z=v_x\tan \alpha =v_1\sin [(\frac{\pi }{2}-\arcsin \frac{v_0}{v_1})\frac{x}{l}+\arcsin \frac{v_0}{v_1}]·\frac{h{\prime }_x}{h_x}z}\end{array}$$

(5)

where, $v_x$ , $v_y$ , and $v_z$ are the velocity components of rolling direction, spreading direction, and compression direction respectively.

According to equation (5), the strain rate field can be obtained from the geometric equation as follows

$$\{\begin{array}{c}{\stackrel{·}{\epsilon }}_x=\frac{\mathrm{d}{v}_x}{\mathrm{d}x}=\frac{v_1(\pi /2-\arcsin v_0/v_1)}{l}\cos \hfill \\ \multicolumn{1}{c}{[(\frac{\pi }{2}-\arcsin v_0/v_1)\frac{x}{l}+\arcsin v_0/v_1]={\stackrel{·}{\epsilon }}_{max}}\\ \multicolumn{1}{c}{{\stackrel{·}{\epsilon }}_y=0}\\ \multicolumn{1}{c}{{\stackrel{·}{\epsilon }}_z=-{\stackrel{·}{\epsilon }}_{max}=-\frac{v_1(\pi /2-\arcsin v_0/v_1)}{l}\cos }\\ \multicolumn{1}{c}{[(\frac{\pi }{2}-\arcsin v_0/v_1)\frac{x}{l}+\arcsin v_0/v_1]={\stackrel{·}{\epsilon }}_{min}}\\ \multicolumn{1}{c}{{\stackrel{·}{\epsilon }}_{\mathrm{xy}}=0}\\ \multicolumn{1}{c}{{\stackrel{·}{\epsilon }}_{\mathrm{yz}}=0}\\ \multicolumn{1}{c}{{\stackrel{·}{\epsilon }}_{\mathrm{zx}}=\frac{1}{2}(\frac{\mathrm{d}{v}_x}{\mathrm{d}z}+\frac{\mathrm{d}{v}_z}{\mathrm{d}x})=\frac{z}{2}[{\stackrel{·}{\epsilon }}_x\frac{h{\prime }_x}{h_x}+v_x\frac{h\prime {\prime }_x}{h_x}-v_x{\left(\frac{h{\prime }_x}{h_x}\right)}^2]}\end{array}$$

(6)

In equations (5) and (6), there are $x=0,$ $v_x=v_0;$ $x=l,$ $v_x=v_1,$ $z=0,$ $v_z=0;$ $z=h_x,$ $v_z=v_x\tan \alpha ;$ ${\stackrel{·}{\epsilon }}_x+{\stackrel{·}{\epsilon }}_y+{\stackrel{·}{\epsilon }}_z=0.$ Therefore, the sinusoidal velocity field satisfies the kinematically admissible condition and can be used for rolling energy analysis.

Internal power of deformation

The internal deformation power during rolling can be calculated by ²⁰

$$W_{\mathrm{i}}=\sqrt{\frac{2}{3}}{\sigma }_{\mathrm{s}}\int \int {\int }_V\sqrt{{\stackrel{·}{\epsilon }}_{\mathrm{ik}}{\stackrel{·}{\epsilon }}_{\mathrm{ik}}}\mathrm{d}V$$

(7)

where ${\sigma }_{\mathrm{s}}$ is the deformation resistance of metal, ${\stackrel{·}{\epsilon }}_{\mathrm{ik}}$ is the strain rate (i =  $x,y,z$ , $k=x,y,z$ ).

Substitute the strain rate field in equation (6) into equation (7), it produces

$$\begin{array}{c}\multicolumn{1}{c}{W_{\mathrm{i}}=4\sqrt{\frac{2}{3}}{\sigma }_{\mathrm{s}}b\int {\int }_S\sqrt{{\stackrel{·}{\epsilon }}_x^2+{\stackrel{·}{\epsilon }}_z^2+{\stackrel{·}{\epsilon }}_{\mathrm{xz}}^2+{\stackrel{·}{\epsilon }}_{\mathrm{zx}}^2}\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8{\sigma }_{\mathrm{s}}}{\sqrt{3}}b\int {\int }_S\sqrt{{\stackrel{·}{\epsilon }}_x^2+{\stackrel{·}{\epsilon }}_{\mathrm{zx}}^2}\mathrm{d}x\mathrm{d}z}\end{array}$$

(8)

where b is the width of the ultra-heavy plate.

Because of the nonlinear specific plastic power of the Mises criterion, equation (8) is difficult to be solved out. Here, a new method called the root vector decomposition method is used to solve this problem. By the root vector decomposition method, equation (8) can be simplified as

$$\begin{array}{c}\multicolumn{1}{c}{W_{\mathrm{i}}=\frac{8{\sigma }_{\mathrm{s}}}{\sqrt{3}}b\int {\int }_S\sqrt{{\stackrel{·}{\epsilon }}_x^2+{\stackrel{·}{\epsilon }}_{\mathrm{zx}}^2}\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8{\sigma }_{\mathrm{s}}}{\sqrt{3}}b{\int }_0^l{\int }_0^{h_x}(\frac{{\stackrel{·}{\epsilon }}_x}{\sqrt{1+{({\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x)}^2}}+\frac{{\stackrel{·}{\epsilon }}_{\mathrm{xz}}}{\sqrt{1+{({\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x)}^{-2}}})}\\ \multicolumn{1}{c}{\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8{\sigma }_{\mathrm{s}}}{\sqrt{3}}b{\int }_0^l{\int }_0^{h_x}({\stackrel{·}{\epsilon }}_x{l}_1+{\stackrel{·}{\epsilon }}_{\mathrm{xz}}{l}_2)\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8{\sigma }_{\mathrm{s}}}{\sqrt{3}}b\int {\int }_S\stackrel{·}{\mathbf{\epsilon }}·{\stackrel{·}{\mathbf{\epsilon }}}^0\mathrm{d}x\mathrm{d}z}\end{array}$$

(9)

where, the strain rate vector and the direction cosine in equation (9) are as follows

$$\stackrel{·}{\mathbf{\epsilon }}={\stackrel{·}{\epsilon }}_x{\mathbf{e}}_1+{\stackrel{·}{\epsilon }}_{\mathrm{xz}}{\mathbf{e}}_2,{\stackrel{·}{\mathbf{\epsilon }}}^0=l_1{\mathbf{e}}_1+l_2{\mathbf{e}}_2$$

(10)

$$l_1={[1+{({\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x)}^2]}^{-1/2},l_2={[1+{({\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x)}^{-2}]}^{-1/2}$$

(11)

The ${\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x$ in equation (9) is calculated by the mean value theorems for definite integrals

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{\stackrel{·}{\epsilon }}_{\mathrm{xz}}}{{\stackrel{·}{\epsilon }}_x}\approx \overline{{\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x}=\frac{z}{2}[\frac{1}{l}{\int }_0^l\frac{h_x^{\prime }}{h_x}\mathrm{d}x+\frac{1}{l}{\int }_0^l\frac{v_x(h_x^{″}{h}_x-h{\prime }_x^2)}{v_x^{\prime }{h}_x^2}\mathrm{d}x]}\\ \multicolumn{1}{c}{=\frac{v_0\tan \theta }{2(v_1-v_0)h_0}z}\end{array}$$

(12)

Denote the pass reduction coefficient by $\eta =h_0/h_1$ in the rolling process, the following formula can be obtained according to the constant volume condition $v_0{h}_{0}b=v_1{h}_{1}b$

$$\frac{v_1}{v_0}=\frac{h_0}{h_1}=\eta$$

(13)

Since $\tan \theta ={-h{\prime }_x|}_{x=l}\approx -2\mathrm{\Delta }h/l$ , equation (12) can be written as

$$\overline{{\stackrel{·}{\epsilon }}_{\mathrm{xz}}/{\stackrel{·}{\epsilon }}_x}=\frac{-(\frac{h_0}{h_1}-1)z}{2(\frac{v_1}{v_0}-1)\eta l}=\frac{-2(\eta -1)z}{2(\eta -1)\eta l}=-\frac{z}{l\eta }$$

(14)

Substitute equation (14) into equation (9), then we can obtain

$$\begin{array}{c}\multicolumn{1}{c}{W_{\mathrm{i}}=I_1+I_2}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta [v_1\ln (\frac{h_1}{l\eta }+\sqrt{1+{\left(\frac{h_1}{l\eta }\right)}^2})-v_0\ln (\frac{h_0}{l\eta }+\sqrt{1+{\left(\frac{h_0}{l\eta }\right)}^2})-\frac{U}{b{h}_m}\ln \frac{h_1+\sqrt{{\left(l\eta \right)}^2+h_1^2}}{h_0+\sqrt{{\left(l\eta \right)}^2+h_0^2}}+v_0\frac{l\eta \tan \theta }{4h_0}(\frac{h_0\sqrt{{\left(l\eta \right)}^2+h_0^2}}{{\left(l\eta \right)}^2}+\frac{1}{2}\ln \frac{\sqrt{{\left(l\eta \right)}^2+h_0^2}-h_0}{\sqrt{{\left(l\eta \right)}^2+h_0^2}+h_0})+\frac{U\ln \eta }{2l\eta \mathrm{bl}}(\sqrt{{\left(l\eta \right)}^2+h_1^2}-\sqrt{{\left(l\eta \right)}^2+h_0^2})]}\end{array}$$

(15)

where $U=v_0{h}_{0}b=v_1{h}_{1}b=v_n{h}_{n}b=v_{\mathrm{R}}b\cos {\alpha }_n(R+h_1-R\cos {\alpha }_n)$ is the volume flow rate, and the derivation of $I_1$ and $I_2$ is as follows

$$\begin{array}{c}\multicolumn{1}{c}{I_1=\frac{8}{\sqrt{3}}{\sigma }_{\mathrm{s}}b{\int }_0^l{\int }_0^{h_x}{\stackrel{·}{\epsilon }}_x{l}_1\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_{s}b{\int }_0^l{\int }_0^{h_x}\frac{{\stackrel{·}{\epsilon }}_x}{\sqrt{1+{(z/l\eta )}^2}}\mathrm{d}x\mathrm{d}z}\end{array}$$

$$\mathrm{set}z/l\eta =\tan \phi ,$$

$$\begin{array}{c}\multicolumn{1}{c}{I_1=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta {\int }_0^l{\int }_0^{\arctan h_x/l\eta }{\stackrel{·}{\epsilon }}_x\sec \phi \mathrm{d}\phi \mathrm{d}x}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta {\int }_0^l\ln (\frac{h_x}{l\eta }+\sqrt{1+{\left(\frac{h_x}{l\eta }\right)}^2}){\stackrel{·}{\epsilon }}_x\mathrm{d}x}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta {\int }_0^l\ln (\frac{h_x}{l\eta }+\sqrt{1+{\left(\frac{h_x}{l\eta }\right)}^2})\mathrm{d}{v}_x}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta \{\left[v_x\ln (\frac{h_x}{l\eta }+\sqrt{1+{\left(\frac{h_x}{l\eta }\right)}^2})\right]{|}_0^l}\\ \multicolumn{1}{c}{-{\int }_0^l{v}_x\mathrm{d}\left[\ln (\frac{h_x}{l\eta }+\sqrt{1+{\left(\frac{h_x}{l\eta }\right)}^2})\right]\}}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_s\mathrm{bl}\eta [v_1\ln (\frac{h_1}{l\eta }+\sqrt{1+{\left(\frac{h_1}{l\eta }\right)}^2})}\\ \multicolumn{1}{c}{-v_0\ln (\frac{h_0}{l\eta }+\sqrt{1+{\left(\frac{h_0}{l\eta }\right)}^2})}\\ \multicolumn{1}{c}{-\frac{U}{{\mathrm{bh}}_m}\ln \frac{h_1+\sqrt{{\left(l\eta \right)}^2+h_1^2}}{h_0+\sqrt{{\left(l\eta \right)}^2+h_0^2}}]}\end{array}$$

$$\begin{array}{c}\multicolumn{1}{c}{I_2=\frac{8}{\sqrt{3}}{\sigma }_{\mathrm{s}}b{\int }_0^l{\int }_0^{h_x}{\stackrel{·}{\epsilon }}_{\mathrm{xz}}{l}_2\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_{s}b{\int }_0^l{\int }_0^{h_x}\frac{{\stackrel{·}{\epsilon }}_{\mathrm{xz}}}{\sqrt{1+{(z/l\eta )}^{-2}}}\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_{s}b{\int }_0^l{\int }_0^{h_x}{\stackrel{·}{\epsilon }}_{\mathrm{xz}}·\sqrt{\frac{z^2}{z^2+{\left(l\eta \right)}^2}}\mathrm{d}x\mathrm{d}z}\\ \multicolumn{1}{c}{=\frac{8}{\sqrt{3}}{\sigma }_{s}b{\int }_0^l{\int }_0^{h_x}[{\stackrel{·}{\epsilon }}_x\frac{h_x^{\prime }}{h_x}+v_x\frac{h_x{h}_x^{″}-{\left(h_x^{\prime }\right)}^2}{h_x^2}]·\frac{z}{2}}\\ \multicolumn{1}{c}{·\sqrt{\frac{z^2}{z^2}+{\left(l\eta \right)}^2}\mathrm{d}x\mathrm{d}z}\end{array}$$

$$\mathrm{set}\sqrt{\frac{z^2}{z^2+{\left(l\eta \right)}^2}}=t$$

$$\begin{array}{c}{I}_2=\frac{8}{\sqrt{3}}{\sigma }_{s}b\cdot \frac{1}{4}{\int }_0^l\mathrm{d}\left(v_x\frac{h{’}_x}{hx}\right){\int }_0^{\sqrt{\frac{h_x^2}{{\left(l\eta \right)}^2+h_x^2}}}t\mathrm{d}\frac{{\left(l\eta t\right)}^2}{1-t^2}\\ =\frac{8}{\sqrt{3}}{\sigma }_{s}b\cdot \frac{{\left(l\eta \right)}^2}{4}{\int }_0^l\mathrm{d}\left(v_x\frac{h{’}_x}{hx}\right){\int }_0^{\sqrt{\frac{h_x^2}{{\left(l\eta \right)}^2+h_x^2}}}t\mathrm{d}\left(\frac{1}{1-t^2}\right)\\ =\frac{8}{\sqrt{3}}{\sigma }_{s}b\cdot \frac{{\left(l\eta \right)}^2}{4}{\int }_0^l\left[\frac{h_x\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_x^2}}{{\left(l\eta \right)}^2}+\frac{1}{2}\ln \frac{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_x^2}-h_x}{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_x^2}+h_x}\right]\mathrm{d}\left(v_x\frac{h{’}_x}{hx}\right)\\ =\frac{8}{\sqrt{3}}{\sigma }_{s}b\cdot \frac{{\left(l\eta \right)}^2}{4}\\ \cdot [v_0\frac{\tan \theta }{h_0}\left(\frac{h_0\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}}{{\left(l\eta \right)}^2}+\frac{1}{2}\ln \frac{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}-h_0}{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}+h_0}\right)+\frac{2U\ln \eta }{{\left(l\eta \right)}^{2}bl}\left(\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_1^2}-\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}\right)]\end{array}$$

Friction power

The formula of the friction power is ²⁰

$$W_{\mathrm{f}}=4{\int }_0^l{\int }_0^{b_x}{\tau }_{\mathrm{f}}\left|\mathrm{\Delta }{v}_{\mathrm{f}}\right|\mathrm{d}s$$

(16)

Due to the co-linear characteristic, the expression can be also written in vector form, that is

$$W_{\mathrm{f}}=4{\int }_0^l{\int }_0^b{\mathbf{\tau }}_{\mathrm{f}}·\mathrm{\Delta }{\mathbf{v}}_{\mathrm{f}}\mathrm{d}s$$

(17)

where, ${\tau }_{\mathrm{f}}=\mathrm{mk}$ is the friction stress, and $k={\sigma }_{\mathrm{s}}/\sqrt{3}$ is the shear yield strength, the tangent velocity discontinuity $\mathbf{\Delta }{\mathbf{v}}_{\mathbf{f}}$ of the roll surface can be written in vector form as

$$\begin{array}{c}\multicolumn{1}{c}{\mathbf{\Delta }{\mathbf{v}}_{\mathbf{f}}=\mathrm{\Delta }{v}_x\mathbf{i}+\mathrm{\Delta }{v}_y\mathbf{j}+\mathrm{\Delta }{v}_z\mathbf{k}=(v_{\mathrm{R}}\cos \alpha -v_x)\mathbf{i}}\\ \multicolumn{1}{c}{+(v_{\mathrm{R}}\sin \alpha -v_x\tan \alpha )\mathbf{k}}\end{array}$$

(18)

According to the equation of the roll surface, the equation of the surface element of the roll surface, and the direction-cosine along the tangent direction are respectively.

$$\mathrm{d}s=\sqrt{1+{\left(h_x^{\prime }\right)}^2}\mathrm{d}x\mathrm{d}y$$

(19)

$$\{\begin{array}{c}\cos \alpha =\pm \sqrt{R^2-{(l-x)}^2}/R\hfill \\ \multicolumn{1}{c}{\cos \beta =0}\\ \multicolumn{1}{c}{\cos \gamma =\pm (l-x)/R=\sin \alpha }\end{array}$$

(20)

Substitute the above conditions into equation (18), and it results in

$$\begin{array}{c}\multicolumn{1}{c}{W_{\mathrm{f}}=4\mathrm{mk}{\int }_0^l{\int }_0^b\cos \alpha (v_{\mathrm{R}}\cos \alpha -v_x)\sec \alpha \mathrm{d}x\mathrm{d}y+}\\ \multicolumn{1}{c}{4\mathrm{mk}{\int }_0^l{\int }_0^b\sin \alpha (v_{\mathrm{R}}\sin \alpha -v_x\tan \alpha )\sec \alpha \mathrm{d}x\mathrm{d}y}\\ \multicolumn{1}{c}{=4\mathrm{mkb}[v_{\mathrm{R}}R(\theta -2a_n)+\frac{\mathrm{UR}}{b{h}_m}\ln \frac{{\tan }^2(\frac{\pi }{4}+\frac{a_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}]}\end{array}$$

(21)

Shear power

The formula of the shear power can be calculated by ²⁰

$$W_{\mathrm{s}}={\int }_{S}k\left|\mathrm{\Delta }v\right|\mathrm{d}S$$

(22)

On the outlet section of the rolling deformation zone, there are $x=0$ , $h{\prime }_x=b{\prime }_x=0$ , ${v_y|}_{x=l}={v_z|}_{x=l}=0$ . Thus, the shear power is only related to the entrance of the deformation zone. Therefore, the total shear power consumed in the rolling process is equal to the shear power consumed on the inlet section. Therefore

$$\begin{array}{c}\multicolumn{1}{c}{W_{\mathrm{s}}=4\mathrm{kb}{\int }_0^{h_0}\left|\mathrm{\Delta }{\overline{v}}_z\right|\mathrm{d}z}\\ \multicolumn{1}{c}{=\mathrm{kU}\tan \theta }\end{array}$$

(23)

Total power functional and its variation

Since the total power functional is $\Phi =W_{\mathrm{i}}+W_{\mathrm{f}}+W_{\mathrm{s}}$ , we have

$$\begin{array}{c}\Phi =8kbl\eta \\ [v_1\ln \left(\frac{h_1}{l\eta }+\sqrt{1+{\left(\frac{h_1}{l\eta }\right)}^2}\right)-v_0\ln \left(\frac{h_0}{l\eta }+\sqrt{1+{\left(\frac{h_0}{l\eta }\right)}^2}\right)-\frac{U}{b{h}_m}\ln \frac{h_1+\sqrt{{\left(l\eta \right)}^2+h_1^2}}{h_0+\sqrt{{\left(l\eta \right)}^2+h_0^2}}\\ +v_0\frac{l\eta \tan \theta }{4h_0}(\frac{h_0\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}}{{\left(l\eta \right)}^2}+\frac{1}{2}\ln \frac{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}-h_0}{\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}+h_0})+\frac{U\ln \eta }{2l\eta bl}\left(\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_1^2}-\sqrt{{\left(l\eta \right)}^2\mathrm{+}{h}_0^2}\right)]\\ +4mkb\left[v_{\mathrm{R}}R\left(\theta -2{\alpha }_n\right)+\frac{UR}{b{h}_m}\ln \frac{{\tan }^2\left(\frac{\mathrm{\pi }}{4}+\frac{{\alpha }_n}{2}\right)}{\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\theta }{2}\right)}\right]+kU\tan \theta \end{array}$$

(24)

According to the upper bound theorem and considering that ${\alpha }_n$ is a pseudo-independent parameter, it can be known that there is an extremum of $\Phi$ at a given ${\alpha }_n$ . The derivative of the total power functional $\Phi$ with respect to ${\alpha }_n$ is

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\partial \Phi }{\partial {\alpha }_n}=\frac{\partial W_{\mathrm{i}}}{\partial {\alpha }_n}+\frac{\partial W_{\mathrm{f}}}{\partial {\alpha }_n}+\frac{\partial W_{\mathrm{s}}}{\partial {\alpha }_n}}\\ \multicolumn{1}{c}{=8\mathrm{kbl}\eta (-\frac{N}{b{h}_m}\ln \frac{h_1+\sqrt{{\left(l\eta \right)}^2+h_1^2}}{h_0+\sqrt{{\left(l\eta \right)}^2+h_0^2}}+\frac{N\ln \eta }{2l\eta \mathrm{bl}}(\sqrt{{\left(l\eta \right)}^2+h_1^2}-\sqrt{{\left(l\eta \right)}^2+h_0^2}))}\\ \multicolumn{1}{c}{+4\mathrm{mk}[\frac{\mathrm{NR}}{h_m}\ln \frac{{\tan }^2(\frac{\pi }{4}+\frac{{\alpha }_n}{2})}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}+\frac{2\mathrm{UR}}{h_m\cos a_n}-2v_R\mathrm{Rb}]+\mathrm{kN}\tan \theta }\end{array}$$

(25)

$$\mathrm{where}N=\partial U/\partial {\alpha }_n=v_{\mathrm{R}}b\sin 2{\alpha }_n-v_{\mathrm{R}}b(R+h_1)\sin {\alpha }_n$$

By substituting the minimum value ${\Phi }_{min}$ of the total power functional determined by the above equation into the following equation, the analytical solutions of rolling torque, rolling force and stress state coefficient can be obtained as follows ²¹

$$\{\begin{array}{c}{M}_{min}=\frac{R}{2v_{\mathrm{R}}}{\Phi }_{min}\hfill \\ \multicolumn{1}{c}{F_{min}=\frac{M_{min}}{\chi \sqrt{2R\mathrm{\Delta }h}}}\\ \multicolumn{1}{c}{n_{\sigma }=\frac{F_{min}}{4\mathrm{blk}}}\end{array}$$

(26)

where $\chi$ represents the force arm coefficient, which is about 0.5 in the hot rolling process.

Temperature distribution along the thickness direction

In the process of gradient temperature rolling, the temperature distribution along the plate thickness direction is not uniform, and there is a big temperature difference between the surface and the core of the plate. Qiu et al. ²² assumed that the temperature distribution from the center to the surface of steel plate in the rolling process conforms to the quadratic temperature model. The temperature distribution equation is assumed to be $T=a{z}^2+\mathrm{bz}+c$ . The boundary conditions are $z=0$ , $T=T_{\mathrm{s}}$ , $T\prime =0$ ; $z=h_0$ , $T=T_{\mathrm{f}}$ , Therefore, the temperature distribution expression can be obtained as

$$T=\left(\frac{T_{\mathrm{f}}-T_{\mathrm{s}}}{h_0^2}\right)z^2+T_{\mathrm{s}}$$

(27)

Based on this, the average temperature can be determined by the integral mean value theorem, expressed as $\overline{T}=(T_{\mathrm{f}}+2T_{\mathrm{s}})/3$ .

The derivative of the above expression gives an expression for the temperature gradient as follows

$$T^{\prime }=2\left(\frac{T_{\mathrm{f}}-T_{\mathrm{s}}}{h_0^2}\right)z$$

(28)

As can be seen from the above equation, the temperature gradient is closely related to surface temperature $T_{\mathrm{f}}$ , core temperature $T_{\mathrm{s}}$ and initial plate thickness $h_0$ , and the position z.

Model validation and parameter analysis

Validation of rolling force model

The measured data from Ding et al. ¹⁶ were used to verify the prediction accuracy of the rolling force model. The rolled steel plate is hot rolled E355DD steel, and the size of the steel plate is $300\mathrm{mm}\times 1800\mathrm{mm}\times 2800\mathrm{mm}$ . The deformation resistance model of the E355DD steel adopted in this paper is ²³

$$\begin{array}{c}\multicolumn{1}{c}{\sigma =167.639\times \exp (-2.079\times \frac{T}{1000}+2.668)}\\ \multicolumn{1}{c}{\times {\left(\frac{\stackrel{·}{\epsilon }}{10}\right)}^{0.299\times \frac{T}{1000}-0.248}}\\ \multicolumn{1}{c}{\times [1.646\times {\left(\frac{\epsilon }{0.4}\right)}^{0.470}-0.646\times \left(\frac{\epsilon }{0.4}\right)]}\end{array}$$

(29)

where T is deformation temperature, $\epsilon$ is strain, $\stackrel{·}{\epsilon }$ is strain rate.

The force arm coefficient in the rolling process can be obtained by referring to the following equation. ²⁴

$$\chi =0.78+0.017(R/h_1)-0.163\sqrt{R/h_1}$$

(30)

where R is the roller radius, $h_1$ is the outlet thickness of rolled plate.

The relevant rolling process parameters are shown in Table 1

Table 1.

Process parameters of gradient temperature rolling process.

Pass no.	Entrythickness( $h_0$ /mm)	Exit thickness( $h_1$ /mm)	Surface temperature( $T_{\mathrm{f}}$ /oC)	Coretemperature( $T_{\mathrm{s}}$ /oC)	Rollingvelocity( $v_0$ / $\mathrm{m}·{\mathrm{s}}^{-1}$ )	The force armcoefficient ( $\chi$ )
1	300.00	273.20	826.3	1149.7	1.00	0.54
2	273.20	247.83	926.8	1149.6	1.00	0.53
3	247.83	226.42	821.1	1148.7	1.00	0.52
4	226.42	207.69	892.1	1147.7	1.00	0.51
5	207.69	179.34	821.5	1144.7	1.00	0.49

Comparison of the analytical results and experimented ones of the rolling force of the above passes is shown in the Table 2

Table 2.

The results of analytical solution of rolling force and the measured ones.

Pass no.	Averagetemperature( $\overline{T}$ /oC)	Measuredrolling force(F/ $\mathrm{kN}$ )	Calculaterolling force( $F_{\mathrm{A}}$ / $\mathrm{kN}$ )	Relative error( $\mathrm{\Delta }=\frac{F_{\mathrm{A}}-F}{F}$ /%)
1	1041.90	29,509,680	30,812,054	4.41
2	1074.33	27,184,320	27,282,236	0.36
3	1039.50	29,184,720	27,574,039	−5.52
4	1062.50	26,318,520	24,157,768	−8.21
5	1036.97	28,920,480	31,873,770	10.21

It can be seen from the Table 2 that the maximum error of analytical rolling force is 10.21%, which is less than the allowable 15% requirement in actual production.

Parameter analysis

Figure 2 shows the deformation resistance of the upper part of the ultra-heavy plate along thickness direction under different surface temperatures. It can be seen from the figure that the lower the surface temperature is, the harder the surface metal of the ultra-heavy plate is, and the resistance of the plate to deformation increases, so the deformation resistance increases. When the surface temperature is the same, the bigger the temperature difference, the average temperature of the whole sheet decreases, the ability to resist deformation increases, and thus the deformation resistance increases.

Figure 2.

Deformation resistance of the ultra-heavy plate along thickness direction under different surface temperatures.

Figure 3 shows the effect of surface temperature and temperature difference between core and surface on rolling force. As shown in the figure, the decrease of the surface temperature of the ultra-heavy plate causes the deformation resistance to increase gradually. With the increase of deformation resistance, the rolling force increases. At the same time, the increase of temperature difference of the ultra-heavy plate will also lead to the increase of its overall deformation resistance. The increase of deformation resistance increases the rolling force. Therefore, it can be also concluded that the surface temperature and the temperature difference are two important parameters that can be used to control the rolling process since they have affected the rolling force evidently.

Figure 3.

The effect of surface temperature and temperature difference on rolling force.

Figure 4 shows the influence of thickness to diameter ratio and reduction ratio on rolling force. It can be seen from the figure that the rolling force gradually increases with the increase of reduction ratio. Under the same reduction rate, the rolling force increases with the thickness to diameter ratio. The change trend indicates that for a given roller, a plate with a bigger thickness or exerted by a bigger reduction rate, the required rolling force for accomplishing the rolling process should become bigger.

Figure 4.

The influence of thickness to diameter ratio and reduction ratio on rolling force.

Figure 5 shows the relationship between shape factor and friction factor on stress state coefficient. It can be seen from the figure that the stress state coefficient decreases with the increase of the shape factor, and the curve slope becomes smaller and smaller. This indicates that a bigger value of shape factor is advocated since it is in favor of improving the stability of the rolling process. The increase of friction factor will lead to the decrease of stress state coefficient, but the effect is not obvious, which indicates that the friction factor is not the key parameter that should be controlled elaborately in the actual production.

Figure 5.

Relationship between shape factor and friction factor on stress state coefficient.

Conclusions

1. A two-dimensional sinusoidal velocity field satisfying the kinematically admissible condition is proposed, which can be used to analyze the ultra-heavy plate rolling. Then, the expression of internal deformation power is derived by the root vector decomposition method. Also, the friction power and shear power are obtained. Ultimately, the analytical solution of rolling force is obtained. Through data comparison, the maximum error is 10.21%.

2. Through analysis, it is found that the deformation resistance from the surface to the core decreases gradually during the gradient temperature rolling and under the same temperature gradient, the deformation resistance increases with the decrease of surface temperature. The rolling force increases when the surface temperature decreases and the temperature difference increases. The force also increases with the increases of the thickness to diameter ratio and the reduction ratio.

3. The increase of the shape factor and the friction factor will lead to the decrease of the stress state coefficient, and the shape factor plays a dominant role.