CN103203371B - Double closed-loop control method for roll gap position pressure of cold rolling mill - Google Patents

Abstract

The invention relates to a double closed-loop control method for roll gap position pressure of a cold rolling mill in the field of cold-rolled strip steel rolling, which is suitable for a cold rolling mill main pressure in a rolling force control mode. The method comprises the following steps: calculating the asymmetric rolling force on the two sides of the rolling mill after the rolling mill is additionally inclined; according to the Kaplan energy theorem, elastic bending influence functions in the forms of supporting rollers and working roller simple beams suitable for asymmetric rolling calculation are deduced; the roller system deformation theory and the metal transverse flow theory are effectively integrated; the deformation coordination equation in the traditional roller system deformation theory is effectively improved; and (4) in consideration of errors of theoretical calculation, carrying out real-time online self-adaptive correction on the theoretical calculation model. The application of the technology can effectively inhibit the belt breakage accident caused by excessive inclination adjustment, and has remarkable effect on controlling the single-side waves.

Description

Cold-rolling mill Roll-gap position pressure double-closed-loop control method

Technical field

The present invention relates to Cold Rolling field, it is adaptable to cold rolling owner presses as roll-force control model.This technology Application can effectively suppress the generation of broken belt accident that causes because tilt adjustments excess, the control to monolateral wave simultaneously has Significantly effect.

Background technology

Thickness of slab precision and plate shape are two big quality index of decision plate band geometric accuracy.The control of thickness of slab precision, Development through for many years is gradually improved.And Strip Shape Control is complicated and changeable due to influence factor, basic theory, detection technique and The aspects such as control technology also have many problems to be solved the most at all, even to this day, and control skill theoretical about Strip Shape Control The research of art is still constantly being improved and is being updated.

Cold-rolled strip steel shape controls to include: the setting of target flatness, the measurement of plate shape, the process of measured data and plate shape control The adjustment of executing agency processed.To the plate shape actual measurement data collected, a multinomial is generally used to carry out returning and orthogonal Decompose, resolve into once, secondary, three times, four times and high order plate shape component, with the measured value of each component and the deviation of desired value be Adjustment amount, utilizes roller, roll shifting, sub-sectional cooling and roller declination to eliminate various flatness defect.Adjustment bending roller force is permissible Change the convexity of roll gap, eliminate the secondary in plate shape deviation and four component of degree n ns.Roll shifting can change the contact condition of roller system, Eliminate working roll and be harmful to the impact of moment of flexure, improve roller efficiency and reduce the edge thinning of band steel.Cannot by roller declination and Roller controls the high-order component flatness defect eliminated, and available segment cooling is controlled.

Roller declination is by adjusting transmission side and the position realization of fore side support roller hydraulic cylinder, belonging to position control System.After roller declination, constitute wedge shape roll gap, for eliminating once with three asymmetric flatness defects in plate shape offset component. The roll control system of cold-rolling mill is embedded in and presses in control system, and the control mode pressing system has two kinds: position control and Roll-force controls.If the system that presses uses position control mode, after inclination control is embedded, setting value belongs to position quantity Guiding principle, the stability of its control system is mathematical.If as it is shown in figure 1, the system that presses uses roll-force control mode, tilting Control is embedded, and just constitutes position and pressure double closed-loop control system.By hydraulic system oil sources pressure and press cylinder size Limiting, rolling force setup value is without amplitude limit control, and tilts setting value and give the definite value amplitude limit of a 2.0mm.Inclination control Setting value is demarcated tilt quantity, strip tension deviation adjusting amount and plate shape tilt adjustments amount superposition by manual intervention amount, milling train and is constituted.

Different-thickness, different in width, different supplied materials wedge shape and the cold-strip steel of unlike material, simultaneously by edge thickness, limit Portion's quality and the impact of plate shape measurement system accuracy, the 2.0mm fixed value amplitude limit tilting setting value is the most excessive, tilts setting value It is improper slightly to regulate, and may result in band steel broken belt.Fig. 2 is the inclination during actual rolling broken belt, both sides roll-force and rolling The PDA curve of power difference, when tilting setting value and being constant, the roll-force difference of transmission side and fore side is held essentially constant, with The increase of actual value, roll-force difference should continue to increase, but keeps definite value, to strongly reducing later, band is described Steel has occurred that edge cracks, broken belt process start.Analyze reason: one is to tilt setting value to regulate excessive；Two is the edge of band steel Quality existing defects.Judge in the operation of rolling that the PDA curve of whether broken belt is as it is shown on figure 3, broken belt stop sign is passed by rolling mill main Dynamic system sends, it is judged that according to being: mill speed be not 0 and tension force be 0.

Before band steel generation broken belt, have started to that obvious broken belt sign occurs: the inclination of roll has greatly changed, But fore side remains in that constant with the roll-force difference of transmission side.Amplitude limit value owing to tilting is ± 1.0mm, 20s before broken belt In, tilt and reached amplitude limit value, and the difference of roll-force does not changes along with the change tilted.If tilting value at this moment can be limited Continue to increase or stop in time, be then avoided that the generation of broken belt accident.

Summary of the invention

It is an object of the invention to provide a kind of cold-rolling mill Roll-gap position pressure double-closed-loop control method, the application energy of the method The generation of the broken belt accident that enough effectively suppression cause because of tilt adjustments excess, the control to monolateral wave simultaneously has significantly effect Really.

For achieving the above object, the present invention is achieved through the following technical solutions:

Cold-rolling mill Roll-gap position pressure double-closed-loop control method, after milling train additional tilt, bilateral Asymmetric Rolling power calculates； Derive according to Ka Shi energy theorem and be suitable to support roller, the elastic bending shadow of working roll simply supported beam form that Asymmetric Rolling calculates Ring function；Rolls' deformation theory effectively integrates with metal theory of cross-flow；Deformation in conventional roll system Deformation Theory is assisted Equation is adjusted to be effectively improved；The error calculated in view of theory, carries out real-time online self adaptation to theoretical calculation model Revise；

After rolling mill additional tilt, contact in rolling is the most no longer traditional centered by rolling centerline Symmetrical, but asymmetrical, when carrying out asymmetric calculating, the roll-force of the left and right sides is unknown number；List when calculating Unit is divided into and arranges from left to right along roller whole body, and support bending influence function of deriving is:

$g_b(i,j)=\frac{1}{{3E}_b{I}_{b1}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}{l}_O^3+\frac{1}{{3E}_b{I}_{b2}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}(x_j^3-l_O^3)$

$+\frac{1}{E_b{I}_{b2}}\frac{l_b-x_i}{l_b}\{-\frac{x_j}{l_b}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_b{I}_{b2}}\left\{\frac{x_i{x}_j}{l_b^2}\frac{1}{3}\right[l_D^3-x_i^3]-\frac{x_i{x}_j}{l_b}[l_D^2-x_i^2]+x_i{x}_j[l_D-x_i\left]\right\}$

$\begin{array}{cc}+\frac{1}{E_b{I}_{b1}}\left\{\frac{x_i{x}_j}{l_b^2}\frac{1}{3}\right[l_b^3-l_D^3]-\frac{x_i{x}_j}{l_b}[l_b^2-l_D^2]+x_i{x}_j[l_b-l_D\left]\right\}& \left(x_i{>x}_j\right)\end{array}$

$+\frac{\mathrm{\Φ}}{{\mathrm{GA}}_{b1}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}{l}_O+\frac{\mathrm{\Φ}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}(x_j-l_O)$

$+\frac{\mathrm{\Φ}}{{\mathrm{GA}}_{b2}}(-\frac{x_j}{l_b})\frac{l_b-x_i}{l_b}(x_i-x_j)+\frac{\mathrm{\Φ}}{{\mathrm{GA}}_{b2}}(-\frac{x_j}{l_b})\frac{l_b-x_i}{l_b}(l_D-x_i)$

$+\frac{\mathrm{\Φ}}{{\mathrm{GA}}_{b1}}\left(\frac{x_j{x}_i}{l_b^2}\right)(l_b-l_D)$

$g_b(i,j)=\frac{1}{{3E}_b{I}_{b1}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}{l}_O^3+\frac{1}{{3E}_b{I}_{b2}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}(x_i^3-l_O^3)$

$+\frac{1}{E_b{I}_{b2}}\frac{l_b-x_j}{l_b}[\frac{x_i}{2}(x_j^2-x_i^2)-\frac{x_i}{{3l}_b}(x_j^3-x_i^3)]$

$+\frac{1}{E_b{I}_{b2}}[\frac{x_i{x}_j}{l_b^2}\frac{1}{3}(l_D^3-x_j^3)-\frac{x_i{x}_j}{l_b}(l_D^2-x_j^2)+x_i{x}_j(l_D-x_j)]$

$\begin{array}{cc}+\frac{1}{E_b{I}_{b1}}[\frac{x_i{x}_j}{l_b^2}\frac{1}{3}(l_b^3-l_D^3)-\frac{x_i{x}_j}{l_b}(l_b^2-l_D^2)+x_i{x}_j(l_b-l_D)]& \left(x_i{<x}_j\right)\end{array}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}{l}_O+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}(x_i-l_O)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}(-\frac{x_i}{l_b})(x_j-x_i)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\left(\frac{x_j{x}_i}{l_b^2}\right)(l_D-x_j)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\left(\frac{x_j{x}_i}{l_b^2}\right)(l_b-l_D)$

Working roller bending influence function is:

$g_w(i,j)=\frac{1}{{3E}_w{I}_{w1}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}{l}_{\mathrm{OW}}^3+\frac{1}{{3E}_w{I}_{w2}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}(x_j^3-l_{\mathrm{OW}}^3)$

$+\frac{1}{E_w{I}_{w2}}\frac{l_w-x_i}{l_w}\{-\frac{x_j}{l_w}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_w{I}_{w2}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_{\mathrm{DW}}^3-x_i^3]-\frac{x_i{x}_j}{l_w}[l_{\mathrm{DW}}^2-x_i^2]+x_i{x}_j[l_{\mathrm{DW}}-x_i\left]\right\}$

$\begin{array}{cc}+\frac{1}{E_w{I}_{w1}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_w^3-l_{\mathrm{DW}}^3]-\frac{x_i{x}_j}{l_w}[l_w^2-l_{\mathrm{DW}}^2]+x_i{x}_j[l_w-l_{\mathrm{DW}}\left]\right\}& \left(x_i{>x}_j\right)\end{array}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}{l}_{\mathrm{OW}}+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}(x_j-l_{\mathrm{OW}})$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(x_i-x_j)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(l_{\mathrm{DW}}-x_i)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\left(\frac{x_j{x}_i}{l_w^2}\right)(l_w-l_{\mathrm{DW}})$

$g_w(i,j)=\frac{1}{{3E}_w{I}_{w1}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}{l}_{\mathrm{OW}}^3+\frac{1}{{3E}_w{I}_{w2}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}(x_j^3-l_{\mathrm{OW}}^3)$

$+\frac{1}{E_w{I}_{w2}}\frac{l_w-x_i}{l_w}\{-\frac{x_j}{l_w}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_w{I}_{w2}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_{\mathrm{DW}}^3-x_i^3]-\frac{x_i{x}_j}{l_w}[l_{\mathrm{DW}}^2-x_i^2]+x_i{x}_j[l_{\mathrm{DW}}-x_i\left]\right\}$

$\begin{array}{cc}+\frac{1}{E_w{I}_{w1}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_w^3-l_{\mathrm{DW}}^3]-\frac{x_i{x}_j}{l_w}[l_w^2-l_{\mathrm{DW}}^2]+x_i{x}_j[l_w-l_{\mathrm{DW}}\left]\right\}& \left(x_i{<x}_j\right)\end{array}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}{l}_{\mathrm{OW}}+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}(x_j-l_{\mathrm{OW}})$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(x_i-x_j)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(l_{\mathrm{DW}}-x_i)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\left(\frac{x_j{x}_i}{l_w^2}\right)(l_w-l_{\mathrm{DW}})$

Coordinate deformation equation between working roll and support roller is:

In formula

$\stackrel{r}{T}={\left[\begin{array}{cccc}t\left(1\right)& t\left(2\right)& L& t\left(\mathrm{NB}\right)\end{array}\right]}^T$ For supporting the tilt vector of roller；

${\stackrel{r}{Y}}_{\mathrm{wb}}={\left[\begin{array}{cccc}{y}_{\mathrm{wb}}\left(1\right)& y_{\mathrm{wb}}\left(2\right)& L& y_{\mathrm{wb}}\left(n\right)\end{array}\right]}^T$ For flattening vector between roller,

${\stackrel{r}{Y}}_{\mathrm{wb}0}={\left[\begin{array}{cccc}{y}_{\mathrm{wb}0}\left(0\right)& y_{\mathrm{wb}0}\left(0\right)& L& y_{\mathrm{wb}0}\left(0\right)\end{array}\right]}^T$ Constant vector, i.e. the flattening amount of roll surface center；

${\stackrel{r}{M}}_b={\left[\begin{array}{cccc}{m}_b\left(1\right)& m_b\left(2\right)& L& m_b\left(n\right)\end{array}\right]}^T$ It is to support roll crown vector；

${\stackrel{r}{M}}_w={\left[\begin{array}{cccc}{m}_w\left(1\right)& m_w\left(2\right)& L& m_w\left(n\right)\end{array}\right]}^T$ It it is work roll crown vector.

Calculated by rolls' deformation and deformed area metal lateral flow after additional tilt, finally can obtain both sides roll-force and Difference is:

Transmission side roll-force: $F_D=\frac{1}{l_b}\underset{i=1}{\overset{\mathrm{NB}}{\mathrm{\&Sigma;}}}q\left(i\right)X\left(i\right)$

Fore side roll-force is: $F_O=\underset{i=1}{\overset{\mathrm{NB}}{\mathrm{\&Sigma;}}}q\left(i\right)-F_D$

Both sides roll-force difference is: Δ F=F_D-F_O

With roll-force difference as independent variable, to tilt as dependent variable, go out roll-force difference and inclination with fitting of a polynomial Functional relation is

$\mathrm{tilt}=\underset{i=0}{\overset{n}{\mathrm{\&Sigma;}}}{a}_0\left(i\right){\left(\mathrm{\&Delta;F}\right)}^i$

In formula: a₀For fitting coefficient.

For eliminating the error of theoretical calculation model, can be according to the actual roll-force of detection is poor and actual tilt data are to [Δ F_act(k), tilt_act(k)] (k=1,2,3 ...), by least square method, incline according to error sum of squares minimum principle dynamic corrections Each term coefficient in tiltedly-roll-force difference multinomial, after correction, the functional relation of roll-force difference and inclination is

$\mathrm{tilt}=\underset{i=0}{\overset{n}{\mathrm{\&Sigma;}}}{a}_1\left(i\right){\left(\mathrm{\&Delta;F}\right)}^i.$

In formula: a₁For multinomial coefficient after revising.

The invention has the beneficial effects as follows: the application of the method can effectively suppress the broken belt caused because of tilt adjustments excess The generation of accident, the control to monolateral wave simultaneously has significant effect.

Accompanying drawing explanation

Fig. 1 cold-rolling mill position and pressure double closed-loop control system Fig. 2 operation roll of mill symmetry roller hydraulic schematic diagram；

The roll-force of Fig. 2 actual broken belt process and inclination PDA curve；

Fig. 3 actual operation of rolling band of discontinuance signal PDA curve；

Fig. 4 both sides roll-force mathematic interpolation flow process；

Fig. 5 roll rolled piece discrete codes；

Fig. 6 supports roller force diagram (xi ＞ xj)；

Fig. 7 supports roller force diagram (xi ＜ xj)；

Fig. 8 working roll force diagram (xi ＞ xj)；

Fig. 9 working roll force diagram (xi ＜ xj)；

Figure 10 beam type working roll force diagram；

Band steel dividing elements in the calculating of Figure 11 cross direction profiles tension force；

Figure 12 supports roller force analysis；

Figure 13 result of calculation；

Position roll-force double closed-loop control system after Figure 14 improvement.

Detailed description of the invention

A kind of cold-rolling mill Roll-gap position pressure double-closed-loop control method, calculates additional tilt back roller system by influence function method Deformation, rolls by Cubic Spline Functions Fitting simultaneously and carries out tension force correction into the distribution of, exit thickness, finally draw different-thickness, Different in width and unlike material blank after additional tilt bilateral roll-force difference and tilt between relation, use roll-force difference Dynamic corrections tilts the amplitude limit of setting value.

The operation of rolling of Strip is an extremely complex metal-pressed machine process, roll rear Strip strip crown and Plate shape is decided by rolled piece 3 D deformation in roll gap.Metal three-dimensional plastic model provides rolling pressure for rolls' deformation model Power and cross direction profiles thereof, rolls' deformation model provides for metal three-dimensional plastic model and rolls rear thickness of strip cross direction profiles.Cause The 3 D deformation model of this foundation metal and rolls' deformation model, can calculate the relation between inclination and both sides roll-force difference, Concrete calculation process is shown in Fig. 4.

Influence function method is a kind of method of discretization.Its basic thought is, roll is separated into some unit, will Load that roll is born and formed bits for mill roller are also by same unit discretization, about influence function in applied mathematics physics The deformation that concept causes at each point when first determining and each unit is applied unit force, draws in each unit time then by whole load effect The deformation superposition risen, just draws the deflection of each unit, such that it is able to calculate the thickness distribution in exit, in conjunction with metal at roll gap In lateral flow can calculate outlet strip tension distribution.

Being made as example with four rollings herein and calculate the relation of additional tilt and both sides roll-force difference, the method can also be promoted To six-high cluster mill.

1 discretization

Rolled piece and roll discretization process have two kinds of method for numbering serial, and in the 1st, method is to arrange from left to right along roller whole body, the 2 kinds of methods are to be arranged by roll centre two ends to the left and right.Considering the symmetry of rolling, current bibliography both domestic and external all uses Discrete method for numbering serial in 2nd.The particularity applied due to this technology, after additional tilt, contact in rolling is the most no longer with roll line Centered by symmetrical, therefore working roll and support roller mesh discretization coding method use the 1st kind of coding method, see Fig. 5.

2 support roller elastic bending influence function g_b

2.1 support roller elastic bending influence function g_b(xi ＞ xj)

The stress supporting roller can be reduced to simply supported beam form as shown in Figure 6, and influence function can be obtained with Castigliano.? Distance left support point distance is xj point function unit power 1, is xi point effect virtual force in distance left support point distanceAnd xi ＞ Xj, then at the deflection of xi point, i.e. the influence function of xi point is by xj point:

$g_b(i,j)=\frac{1}{{3E}_b{I}_{b1}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}{l}_O^3+\frac{1}{{3E}_b{I}_{b2}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}(x_j^3-l_O^3)$

$+\frac{1}{E_b{I}_{b2}}\frac{l_b-x_i}{l_b}\{-\frac{x_j}{l_b}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_b{I}_{b2}}\left\{\frac{x_i{x}_j}{l_b^2}\frac{1}{3}\right[l_D^3-x_i^3]-\frac{x_i{x}_j}{l_b}[l_D^2-x_i^2]+x_i{x}_j[l_D-x_i\left]\right\}$

$+\frac{1}{E_b{I}_{b1}}\left\{\frac{x_i{x}_j}{l_b^2}\frac{1}{3}\right[l_b^3-l_D^3]-\frac{x_i{x}_j}{l_b}[l_b^2-l_D^2]+x_i{x}_j[l_b-l_D\left]\right\}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}{l}_O+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}(x_j-l_O)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}(-\frac{x_j}{l_b})\frac{l_b-x_i}{l_b}(x_i-x_j)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}(-\frac{x_j}{l_b})\frac{l_b-x_i}{l_b}(l_D-x_i)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\left(\frac{x_j{x}_i}{l_b^2}\right)(l_b-l_D)$

In formula, E_bFor supporting roller elastic modelling quantity；I_b1、I_b2For supporting roll shaft footpath and the bending resistant section coefficient of the body of roll；Φ is for cutting Shear deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_b1For the cross-sectional area at the diameter of axle；A_b2Cross section for the body of roll Long-pending.

2.2 support roller elastic bending influence function g_b(xi ＜ xj)

As it is shown in fig. 7, as xi ＜ xj, be xj point function unit power 1 in distance left support point distance, in distance left support Point distance is xi point effect virtual forceThe influence function of xi point can be obtained by xj point with Castigliano.

$g_b(i,j)=\frac{1}{{3E}_b{I}_{b1}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}{l}_O^3+\frac{1}{{3E}_b{I}_{b2}}\frac{l_b-x_i}{l_b}\frac{l_b-x_j}{l_b}(x_i^3-l_O^3)$

$+\frac{1}{E_b{I}_{b2}}\frac{l_b-x_j}{l_b}[\frac{x_i}{2}(x_j^2-x_i^2)-\frac{x_i}{{3l}_b}(x_j^3-x_i^3)]$

$+\frac{1}{E_b{I}_{b2}}[\frac{x_i{x}_j}{l_b^2}\frac{1}{3}(l_D^3-x_j^3)-\frac{x_i{x}_j}{l_b}(l_D^2-x_j^2)+x_i{x}_j(l_D-x_j)]$

$+\frac{1}{E_b{I}_{b1}}[\frac{x_i{x}_j}{l_b^2}\frac{1}{3}(l_b^3-l_D^3)-\frac{x_i{x}_j}{l_b}(l_b^2-l_D^2)+x_i{x}_j(l_b-l_D)]$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}{l}_O+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}\frac{l_b-x_i}{l_b}(x_j-l_O)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}\frac{l_b-x_j}{l_b}(-\frac{x_i}{l_b})(x_j-x_i)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b2}}(-\frac{x_j{x}_i}{l_b^2})(l_D-x_j)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{b1}}\left(\frac{x_j{x}_i}{l_b^2}\right)(l_b-l_D)$

In formula, E_bFor supporting roller elastic modelling quantity；I_b1、I_b2For supporting roll shaft footpath and the bending resistant section coefficient of the body of roll；Φ is for cutting Shear deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_b1For the cross-sectional area at the diameter of axle；A_b2Cross section for the body of roll Long-pending.

Support bending influence function is write as matrix form and is

3 working roll elastic bending influence function g_w

3.1 working roll elastic bending influence function g_w(xi ＞ xj)

As shown in Figure 8, as xi ＞ xj, it is xj point function unit power 1 in distance left support point distance, in distance left support Point distance is xi point effect virtual forceThe influence function of xi point can be obtained by xj point with Castigliano.

$g_w(i,j)=\frac{1}{{3E}_w{I}_{w1}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}{l}_{\mathrm{OW}}^3+\frac{1}{{3E}_w{I}_{w2}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}(x_j^3-l_{\mathrm{OW}}^3)$

$+\frac{1}{E_w{I}_{w2}}\frac{l_w-x_i}{l_w}\{-\frac{x_j}{l_w}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_w{I}_{w2}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_{\mathrm{DW}}^3-x_i^3]-\frac{x_i{x}_j}{l_w}[l_{\mathrm{DW}}^2-x_i^2]+x_i{x}_j[l_{\mathrm{DW}}-x_i\left]\right\}$

$+\frac{1}{E_w{I}_{w1}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_w^3-l_{\mathrm{DW}}^3]-\frac{x_i{x}_j}{l_w}[l_w^2-l_{\mathrm{DW}}^2]+x_i{x}_j[l_w-l_{\mathrm{DW}}\left]\right\}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}{l}_{\mathrm{OW}}+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}(x_j-l_{\mathrm{OW}})$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(x_i-x_j)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(l_{\mathrm{DW}}-x_i)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\left(\frac{x_j{x}_i}{l_w^2}\right)(l_w-l_{\mathrm{DW}})$

In formula, E_wFor working roll elastic modelling quantity；I_w1、I_w2Bending resistant section coefficient for the working roll diameter of axle and the body of roll；Φ is for cutting Shear deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_w1For the cross-sectional area at the working roll diameter of axle；A_w2For working roll The cross-sectional area of the body of roll.

3.2 working roll elastic bending influence function g_w(xi ＜ xj)

As it is shown in figure 9, as xi ＜ xj, be xj point function unit power 1 in distance left support point distance, in distance left support Point distance is xi point effect virtual forceThe influence function of xi point can be obtained by xj point with Castigliano.

$g_w(i,j)=\frac{1}{{3E}_w{I}_{w1}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}{l}_{\mathrm{OW}}^3+\frac{1}{{3E}_w{I}_{w2}}\frac{l_w-x_i}{l_w}\frac{l_w-x_j}{l_w}(x_j^3-l_{\mathrm{OW}}^3)$

$+\frac{1}{E_w{I}_{w2}}\frac{l_w-x_i}{l_w}\{-\frac{x_j}{l_w}\frac{1}{3}[x_i^3-x_j^3]+\frac{1}{2}{x}_j[x_i^2-x_j^2\left]\right\}$

$+\frac{1}{E_w{I}_{w2}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_{\mathrm{DW}}^3-x_i^3]-\frac{x_i{x}_j}{l_w}[l_{\mathrm{DW}}^2-x_i^2]+x_i{x}_j[l_{\mathrm{DW}}-x_i\left]\right\}$

$+\frac{1}{E_w{I}_{w1}}\left\{\frac{x_i{x}_j}{l_w^2}\frac{1}{3}\right[l_w^3-l_{\mathrm{DW}}^3]-\frac{x_i{x}_j}{l_w}[l_w^2-l_{\mathrm{DW}}^2]+x_i{x}_j[l_w-l_{\mathrm{DW}}\left]\right\}$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}{l}_{\mathrm{OW}}+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}\frac{l_w-x_j}{l_w}\frac{l_w-x_i}{l_w}(x_j-l_{\mathrm{OW}})$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(x_i-x_j)+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w2}}(-\frac{x_j}{l_w})\frac{l_w-x_i}{l_w}(l_{\mathrm{DW}}-x_i)$

$+\frac{\mathrm{\&Phi;}}{{\mathrm{GA}}_{w1}}\left(\frac{x_j{x}_i}{l_w^2}\right)(l_w-l_{\mathrm{DW}})$

In formula, E_wFor working roll elastic modelling quantity；I_w1、I_w2Bending resistant section coefficient for the working roll diameter of axle and the body of roll；Φ is for cutting Shear deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_w1For the cross-sectional area at the working roll diameter of axle；A_w2For working roll The cross-sectional area of the body of roll.

Working roll elastic bending influence function is write as matrix form

4 rolled pieces and working roll elastic flattening influence function g_ws

The working roll that the working roll elastic flattening caused by draught pressure between rolled piece and working roll uses pool, family to derive is elastic Flattening influence function formula, after being modified according to middle island revised theory, working roll elastic flattening influence function is

$g_{\mathrm{ws}}(i,j)=\frac{1-v_w^2}{{\mathrm{\&pi;E}}_w\mathrm{\&Delta;x}}\{\ln \frac{\sqrt{l_d^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{l_d^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i-\frac{\mathrm{\&Delta;x}}{2}}$

$+\frac{(X_i+\frac{\mathrm{\&Delta;x}}{2})}{l_d}\ln \frac{\sqrt{l_d^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+l_d}{|X_i+\frac{\mathrm{\&Delta;x}}{2}|}-\frac{(X_i-\frac{\mathrm{\&Delta;x}}{2})}{l_d}\ln \frac{\sqrt{l_d^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+l_d}{|X_i-\frac{\mathrm{\&Delta;x}}{2}|}\}$

$-\frac{1}{2(1-v_w)}\&times;[\frac{X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}}-\frac{X_i-\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}}]$

$-\ln \frac{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}-(X_i-\frac{\mathrm{\&Delta;x}}{2})}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}-(X_i+\frac{\mathrm{\&Delta;x}}{2})}\}$

E in formula_wElastic modelling quantity for working roll；v_wPoisson's ratio for working roll；Δ x is each unit width；l_dFor jth list Unit's rolled piece and working roll contact arc length, can be determined by Xi Qikeke formula, l_dIt is a variable along plate width direction, with each unit Draught pressure is different and different；X_iDistance for jth unit center point to i-th cell central point；R_wFor working roll radius.

The working roll elastic flattening influence function that draught pressure causes is write as matrix form and is

5 work roll bending power influence function Gf

As shown in Figure 10, function unit load at the working roll axle head support at rolling center xj, the work being induced by Make roller elastic deformation, i.e. work roll bending power influence function, can be obtained by Castigliano.

$g_f\left(i\right)=\frac{1}{{2E}_w{I}_{w1}}[x_i^2{x}_j-\frac{1}{3}{x}_i^3]=\frac{1}{{6E}_w{I}_{w1}}[x_i^2(\frac{{3l}_B}{2}-x_i)+\frac{5}{6}(1+\mathrm{\&upsi;})D_w^2{x}_i]$

In formula, E_wFor working roll elastic modelling quantity；I_w1Bending resistant section coefficient for the working roll body of roll；D_wFor work roll diameter.

The matrix form of work roll bending power influence function is

${\stackrel{r}{G}}_f={\left[\begin{array}{cccc}{g}_f\left(1\right)& g_f\left(2\right)& L& g_f\left(\mathrm{NW}\right)\end{array}\right]}^T$

Influence function g is flattened between 6 rollers_wb

Flattening influence function between roller is:

$g_{\mathrm{wb}}^b(i,j)=\frac{1-v_b^2}{{\mathrm{\&pi;E}}_b}.\frac{3}{4\mathrm{b\&Delta;x}}\&times;\{2b\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i-\frac{\mathrm{\&Delta;x}}{2}}$

$+2(X_i+\frac{\mathrm{\&Delta;x}}{2})\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{|X_i+\frac{\mathrm{\&Delta;x}}{2}|}-2(X_i-\frac{\mathrm{\&Delta;x}}{2})\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{|X_i-\frac{\mathrm{\&Delta;x}}{2}|}$

$-\frac{1}{b^2}[-\frac{b}{3}(X_i-\frac{\mathrm{\&Delta;x}}{2})\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+\frac{b}{3}(X_i+\frac{\mathrm{\&Delta;x}}{2})\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}$

$+\frac{2}{3}{b}^3\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}-(X_i-\frac{\mathrm{\&Delta;x}}{2})}{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}-(X_i+\frac{\mathrm{\&Delta;x}}{2})}+\frac{1}{6}{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^3\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}-b}$

$-\frac{1}{6}{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^3\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}-b}\left]\right\}$

$-\frac{1-v_b^2}{{\mathrm{\&pi;E}}_b\mathrm{\&Delta;x}}\left\{\frac{1}{2(1-v_b)}\right[\frac{X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_b^2}}-\frac{X_i-\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_b^2}}]$

$+\ln \frac{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_b^2}-(X_i-\frac{\mathrm{\&Delta;x}}{2})}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_b^2}-(X_i+\frac{\mathrm{\&Delta;x}}{2})}\}$

In formula, v_bFor supporting the Poisson's ratio of roller；E_bElastic modelling quantity for backing roll；B is width of contact region between j unit roller Half, is calculated by Herz formula；X_iDistance for jth unit midpoint to i-th cell midpoint；Δ x is the width of each unit；R_bFor Support the radius of roller.

The flattening influence function of working roll is

$g_{\mathrm{wb}}^w(i,j)=\frac{1-v_w^2}{{\mathrm{\&pi;E}}_w}.\frac{3}{4\mathrm{b\&Delta;x}}\&times;\{2b\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+X_i-\frac{\mathrm{\&Delta;x}}{2}}$

$+2(X_i+\frac{\mathrm{\&Delta;x}}{2})\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{|X_i+\frac{\mathrm{\&Delta;x}}{2}|}-2(X_i-\frac{\mathrm{\&Delta;x}}{2})\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{|X_i-\frac{\mathrm{\&Delta;x}}{2}|}$

$-\frac{1}{b^2}[-\frac{b}{3}(X_i-\frac{\mathrm{\&Delta;x}}{2})\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+\frac{b}{3}(X_i+\frac{\mathrm{\&Delta;x}}{2})\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}$

$+\frac{2}{3}{b}^3\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}-(X_i-\frac{\mathrm{\&Delta;x}}{2})}{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}-(X_i+\frac{\mathrm{\&Delta;x}}{2})}+\frac{1}{6}{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^3\ln \frac{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{\sqrt{b^2+{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2}-b}$

$-\frac{1}{6}{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^3\ln \frac{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}+b}{\sqrt{b^2+{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2}-b}\left]\right\}$

$-\frac{1-v_w^2}{{\mathrm{\&pi;E}}_w\mathrm{\&Delta;x}}\left\{\frac{1}{2(1-v_b)}\right[\frac{X_i+\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}}-\frac{X_i-\frac{\mathrm{\&Delta;x}}{2}}{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}}]$

$+\ln \frac{\sqrt{{(X_i-\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}-(X_i-\frac{\mathrm{\&Delta;x}}{2})}{\sqrt{{(X_i+\frac{\mathrm{\&Delta;x}}{2})}^2+R_w^2}-(X_i+\frac{\mathrm{\&Delta;x}}{2})}\}$

In formula, v_wPoisson's ratio for working roll；E_wElastic modelling quantity for working roll；B is width of contact region between j unit roller Half, is calculated by Herz formula；X_iDistance for jth unit midpoint to i-th cell midpoint；Δ x is the width of each unit；R_wFor The radius of working roll.

Flattening influence function between roller is

$g_{\mathrm{wb}}(i,j)=g_{\mathrm{wb}}^b(i,j)+g_{\mathrm{wb}}^w(i,j)$

The matrix form flattening influence function between roller is

${\stackrel{\mathrm{\&rho;}}{G}}_{\mathrm{wb}}=\left[\begin{array}{ccc}{g}_{\mathrm{wb}}\left(\mathrm{1,1}\right)& \mathrm{\&Lambda;}& g_{\mathrm{wb}}(1,n)\\ M& & M\\ g_{\mathrm{wb}}(n,1)& \mathrm{\&Lambda;}& g_{\mathrm{wb}}(n,n)\end{array}\right]$

N=min{NW in formula, NB}.

7 equations of static equilibrium

The working roll equation of static equilibrium is

In formula,For draught pressure；F_WDBending roller force for working roll transmission side；F_WOBending roller force for working roll fore side； For roll gap pressure； $\stackrel{r}{P}={\left[\begin{array}{cccc}p\left(1\right)& p\left(2\right)& L& p\left(\mathrm{NS}\right)\end{array}\right]}^T,$ $\stackrel{r}{Q}={\left[\begin{array}{cccc}q\left(1\right)& q\left(2\right)& L& q\left(n\right)\end{array}\right]}^T.$

The equation of static equilibrium supporting roller is

FR in formula_WDFor supporting roller transmission side roll-force；FR_WOFor supporting roller fore side roll-force.

8 compatibility of deformation relation equations

8.1 working rolls and the compatibility of deformation relation supported between roller

In formula

$\stackrel{r}{T}={\left[\begin{array}{cccc}t\left(1\right)& t\left(2\right)& L& t\left(\mathrm{NB}\right)\end{array}\right]}^T$ For supporting the tilt vector of roller.

${\stackrel{r}{Y}}_{\mathrm{wb}}={\left[\begin{array}{cccc}{y}_{\mathrm{wb}}\left(1\right)& y_{\mathrm{wb}}\left(2\right)& L& y_{\mathrm{wb}}\left(n\right)\end{array}\right]}^T$ For flattening vector between roller,

${\stackrel{r}{Y}}_{\mathrm{wb}0}={\left[\begin{array}{cccc}{y}_{\mathrm{wb}0}\left(0\right)& y_{\mathrm{wb}0}\left(0\right)& L& y_{\mathrm{wb}0}\left(0\right)\end{array}\right]}^T$ Constant vector, i.e. the flattening amount of roll surface center.

${\stackrel{r}{M}}_b={\left[\begin{array}{cccc}{m}_b\left(1\right)& m_b\left(2\right)& L& m_b\left(n\right)\end{array}\right]}^T$ It is to support roll crown vector.

${\stackrel{r}{M}}_w={\left[\begin{array}{cccc}{m}_w\left(1\right)& m_w\left(2\right)& L& m_w\left(n\right)\end{array}\right]}^T$ It it is work roll crown vector.

Compatibility of deformation relation between 8.2 working rolls and band steel

In formula

In formulaIt it is rolled piece outgoing gauge vector；

Constant vector, i.e. plate center outgoing gauge；

Constant vector, i.e. the flattening amount of plate center；

${\stackrel{r}{Y}}_{\mathrm{ws}}={\left[\begin{array}{cccc}{y}_{\mathrm{ws}}\left(1\right)& y_{\mathrm{ws}}\left(2\right)& L& y_{\mathrm{ws}}\left(\mathrm{NS}\right)\end{array}\right]}^T,$ The working roll elastic flattening caused by draught pressure to Amount,

${\stackrel{r}{Y}}_w={\left[\begin{array}{cccc}{y}_w\left(1\right)& w_w\left(2\right)& L& y_w\left(\mathrm{NW}\right)\end{array}\right]}^T,$ For working roll amount of deflection vector；

9 tensile stress distributions calculate

Cold-rolling process use hightension rolling, tension distribution influence of rolled power be distributed, thus affect roll gap shape and Outlet belt steel thickness distribution.Therefore the distribution of outlet of cold-rolling mill thickness must take into the impact of tension force.Cold-strip steel tension distribution Computational methods have the calculus of variations, bar unit method, boundary element method and experimental technique.In this patent, the calculating of tensile stress uses variation energy Method, it is contemplated that the asymmetry of rolling, the entry and exit thickness of band steel uses Cubic Spline Functions Fitting, do not use based on Symmetric high-order moment matching, finally can try to achieve outlet band steel horizontal tensile stress distribution.

${\mathrm{\&sigma;}}_1\left(y\right)=\stackrel{\&OverBar;}{{\mathrm{\&sigma;}}_1}+\frac{E}{1-{\mathrm{\&upsi;}}^2}[1+\frac{h\left(y\right)}{\stackrel{\&OverBar;}{h}}-\frac{H\left(y\right)}{\stackrel{\&OverBar;}{H}}-\frac{L\left(y\right)}{\stackrel{\&OverBar;}{L}}+u^{\&prime;}\left(y\right)-\frac{\mathrm{\&Delta;b}}{B}]$

In formula, H (y), h (y) enter, export thickness of slab Transverse functions, mm；

E, υ band elastic modelling quantity and Poisson's ratio；

The average front tensile stress of band, MPa；

L (y) supplied materials length Transverse functions, mm；

H, h, L H (y), the mean value of h (y), L (y), mm；

U (y) band outlet lateral displacement function, mm；

Δ b band absolute spread, mm；

As shown in figure 11, during the horizontal tensile stress of band steel calculates, band steel is divided into NS unit, the width of each unit For Δ x.Gateway thickness and the strip steel at entry length Cubic Spline Functions Fitting of each unit band steel are

H (y)=A_jy³+B_jy²+C_jy+D_j

H (y)=F_jy³+G_jy²+M_jy+N_j

L (y)=P_jy³+Q_jy²+R_jy+T_j

According to sex change energy minimum principle in the operation of rolling, the method solving Eulerian equation by variation, jth list can be tried to achieve In unit, the lateral displacement function of metal is

U (j)=C₁ch(Ky)+C₂sh(Ky)+d_jy²+e_jy+f_j

In formula, $C_1=-\frac{{\mathrm{\&xi;}}_1+{\mathrm{\&xi;}}_3+{\mathrm{\&xi;}}_5}{{\mathrm{\&xi;}}_2+{\mathrm{\&xi;}}_4+{\mathrm{\&xi;}}_6}$

${\mathrm{\&xi;}}_1=\frac{{\mathrm{kh}}_c}{\mathrm{\&Delta;h}}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}\left\{\begin{array}{c}[({4d}_j{y}_j+{2e}_j)\mathrm{ch}\left({\mathrm{Ky}}_j\right)-({4d}_j{y}_{j-1}+{2e}_j)\mathrm{ch}\left({\mathrm{Ky}}_{j-1}\right)]\\ -\frac{4}{K}{d}_j[\mathrm{sh}\left({\mathrm{Ky}}_j\right)-\mathrm{sh}\left({\mathrm{Ky}}_{j-1}\right)]\end{array}\right\}$

${\mathrm{\&xi;}}_2=\frac{h_c}{\mathrm{\&Delta;h}}\mathrm{kK}[\mathrm{sh}\left(\mathrm{KB}\right)-\mathrm{KB}]$

${\mathrm{\&xi;}}_3=\left\{\begin{array}{c}\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^2}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[({2d}_j{y}_{j-1}+e_j)\mathrm{ch}\left({\mathrm{Ky}}_{j-1}\right)-({2d}_j{y}_j+e_j)\mathrm{ch}\left({\mathrm{Ky}}_j\right)]\\ +\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2d}_j+f_j{K}^2+e_j{K}^2{y}_j+d_j{K}^2{y}_j^2]\mathrm{sh}\left({\mathrm{Ky}}_j\right)\\ -\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2d}_j+f_j{K}^2+e_j{K}^2{y}_{j-1}+d_j{K}^2{y}_{j-1}^2]\mathrm{sh}\left({\mathrm{Ky}}_{j-1}\right)\end{array}\right\}$

${\mathrm{\&xi;}}_4=\frac{4t}{\mathrm{\&Delta;hlK}}[\mathrm{sh}\left(\mathrm{KB}\right)+\mathrm{KB}]$

${\mathrm{\&xi;}}_6=\frac{E}{1-{\mathrm{\&upsi;}}^2}\frac{K}{2}[\mathrm{sh}\left(\mathrm{KB}\right)-\mathrm{KB}]$

$C_2=-\frac{{\mathrm{\&zeta;}}_2+{\mathrm{\&zeta;}}_3+{\mathrm{\&zeta;}}_5+{\mathrm{\&zeta;}}_7+{\mathrm{\&zeta;}}_8+{\mathrm{\&zeta;}}_9+{\mathrm{\&zeta;}}_{10}+{\mathrm{\&zeta;}}_{11}+{\mathrm{\&zeta;}}_{12}}{{\mathrm{\&zeta;}}_1+{\mathrm{\&zeta;}}_4+{\mathrm{\&zeta;}}_6}$

${\mathrm{\&zeta;}}_1=\frac{{\mathrm{kh}}_{c}K}{\mathrm{\&Delta;h}}[\mathrm{sh}\left(\mathrm{KB}\right)+\mathrm{KB}]$

${\mathrm{\&zeta;}}_2=-2\mathrm{ksh}\left(\frac{\mathrm{KB}}{2}\right)$

${\mathrm{\&zeta;}}_3=\frac{{2\mathrm{kh}}_c}{\mathrm{\&Delta;h}}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}\left\{\right[(e_j+{2d}_j{y}_j)\mathrm{sh}\left({\mathrm{Ky}}_j\right)-(e_j+{2d}_j{y}_{j-1})\mathrm{sh}\left({\mathrm{Ky}}_{j-1}\right)]-\frac{{2d}_j[\mathrm{ch}\left({\mathrm{Ky}}_j\right)-\mathrm{ch}\left({\mathrm{Ky}}_{j-1}\right)]}{K}\}$

${\mathrm{\&zeta;}}_4=\frac{4t}{\mathrm{\&Delta;hlk}}[\mathrm{sh}\left(\mathrm{KB}\right)-\mathrm{KB}]$

${\mathrm{\&zeta;}}_5=\left\{\begin{array}{c}\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2d}_j+f_j{K}^2+e_j{K}^2{y}_j+d_j{K}^2{y}_j^2]\mathrm{ch}\left({\mathrm{Ky}}_j\right)\\ -\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2d}_j+f_j{K}^2+e_j{K}^2{y}_{j-1}+d_j{K}^2{y}_{j-1}^2]\mathrm{ch}\left({\mathrm{Ky}}_{j-1}\right)\\ +\frac{8t}{\mathrm{\&Delta;hl}}\frac{1}{K^2}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}\left\{\right[(e_j+{2d}_j{y}_{j-1})\mathrm{sh}\left({\mathrm{Ky}}_{j-1}\right)-(e_j+{2d}_j{y}_j)\mathrm{sh}\left({\mathrm{Ky}}_j\right)\left]\right\}\end{array}\right\}$

${\mathrm{\&zeta;}}_6=\frac{E}{2B(1-{\mathrm{\&upsi;}}^2)}[B^2{K}^2+\mathrm{BKsh}\left(\mathrm{KB}\right)-{8\mathrm{sh}}^2\left(\frac{\mathrm{KB}}{2}\right)]$

${\mathrm{\&zeta;}}_7=-\frac{E}{1-{\mathrm{\&upsi;}}^2}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{6g}_j+l_j{K}^2+{2d}_j{K}^2+{2h}_j{K}^2{y}_j+{3g}_j{K}^2{y}_j^2]\mathrm{ch}\left({\mathrm{Ky}}_j\right)$

${\mathrm{\&zeta;}}_8=\frac{E}{1-{\mathrm{\&upsi;}}^2}\frac{1}{K^3}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{6g}_j+l_j{K}^2+{2d}_j{K}^2+{2h}_j{K}^2{y}_{j-1}+{3g}_j{K}^2{y}_{j-1}^2]\mathrm{ch}\left({\mathrm{Ky}}_{j-1}\right)$

${\mathrm{\&zeta;}}_9=-\frac{E}{6B(1-{\mathrm{\&upsi;}}^2)}\mathrm{sh}\left(\frac{\mathrm{KB}}{2}\right)\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{12m}_j{y}_j+{12e}_j{y}_j+{12d}_j{y}_j^2+{6l}_j{y}_j^2+{4h}_j{y}_j^3+{3g}_j{y}_j^4]$

${\mathrm{\&zeta;}}_{10}=\frac{E}{6B(1-{\mathrm{\&upsi;}}^2)}\mathrm{sh}\left(\frac{\mathrm{KB}}{2}\right)\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{12m}_j{y}_{j-1}+{12e}_j{y}_{j-1}+{12d}_j{y}_{j-1}^2+{6l}_j{y}_{j-1}^2+{4h}_j{y}_{j-1}^3+{3g}_j{y}_{j-1}^4]$

${\mathrm{\&zeta;}}_{11}=\frac{E}{1-{\mathrm{\&upsi;}}^2}\frac{1}{K^2}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2h}_j+m_{j}K+e_{j}K+{6g}_j{y}_j+{2d}_j{K}^2{y}_j+l_j{K}^2{y}_j+h_j{K}^2{y}_j^2+g_j{K}^2{y}_j^3]\mathrm{sh}\left({\mathrm{Ky}}_j\right)$

${\mathrm{\&zeta;}}_{12}=-\frac{E}{1-{\mathrm{\&upsi;}}^2}\frac{1}{K^2}\underset{j=1}{\overset{\mathrm{NS}}{\mathrm{\&Sigma;}}}[{2h}_j+m_{j}K+e_{j}K+{6g}_j{y}_{j-1}+{2d}_j{K}^2{y}_{j-1}+l_j{K}^2{y}_{j-1}+h_j{K}^2{y}_{j-1}^2+g_j{K}^2{y}_{j-1}^3]\mathrm{sh}\left({\mathrm{Ky}}_{j-1}\right)$

$d_j=-\frac{a_j}{K^2}$

$e_j=-\frac{b_j}{K^2}$

$f_j=-\frac{{2a}_j}{K^4}-\frac{c_j}{K^2}$

$a_j=-\frac{3}{\mathrm{\&xi;}}(\frac{F_j}{h}-\frac{A_j}{H}-\frac{P_j}{L})$

$b_j=-\frac{2}{\mathrm{\&xi;}}(\frac{G_j}{h}-\frac{B_j}{H}-\frac{Q_j}{L})$

$c_j=-\frac{1}{\mathrm{\&xi;}}(\frac{M_j}{h}-\frac{C_j}{H}-\frac{R_j}{L})$

In various, k is shear strain intensity；h_cFor deformation zone steel average thickness；Δ h band steel gateway thickness average value Difference；K is and h_c, k, process that Δ h, E, υ etc. are relevant calculate variable；T is surface, deformed area average friction force；L is roll flattening Arc length；ξ is and υ, E, k, h_c, process relevant for Δ h etc. calculate variable.

10 fore side roll-forces and transmission side roll-force mathematic interpolation

Supporting roller force analysis and see Figure 12, roll is divided into NB unit along axis, and the width of each unit is Δ x, each list Unit's effect concentrfated load q (i), the load of fore side fulcrum is F_o, the load of the transmission side strong point is F_D.According to equalising torque bar Part, each power relative to the fore side strong point and moment be zero, i.e.

$\underset{i=1}{\overset{\mathrm{NB}}{\mathrm{\&Sigma;}}}q\left(i\right)X\left(i\right)-F_D{l}_b=0$

$F_D=\frac{1}{l_b}\underset{i=1}{\overset{\mathrm{NB}}{\mathrm{\&Sigma;}}}q\left(i\right)X\left(i\right)$

$F_O=\underset{i=1}{\overset{\mathrm{NB}}{\mathrm{\&Sigma;}}}q\left(i\right)-F_D$

Both sides roll-force difference is

Δ F=F_D-F_O

Yield strength σ with steel_s=600MPa, strip width is 1380mm, and inlet thickness is 6mm, and reduction ratio is 0.32, Inlet set tension force is 130MPa, and outlet band steel sets tension force 136MPa, supports roller diameter 0.9906m, work roll diameter 0.60007m, work roll crown is 0.02mm, supports roll crown 0.05mm, and each division unit width is 20mm, and bending roller force is 30KN, result of calculation is as shown in figure 13.

Polynomial fitting is

Tilt=3.9 × 10^-10ΔF+2.4×10^-9

11 model adaptation corrections

Influence function method calculates rolls' deformation and energy variation method solves metal lateral flow problem in roll gap, is practical Calculate the distribution of band steel exports transverse gage and the engineering calculating method of tension distribution.But solving rolls' deformation, tension formula Many approximations and simplification has been done so that outlet transverse gage distribution and tensile stress during deriving and solving Euler's differential equation There is certain error in the result of calculation of distribution.Can be according to the inlet thickness of actual band steel, exit thickness, pressure during actual application Amount, resistance of deformation, strip width, outlet set tension force, inlet set tension force, working roll and support roller roller footpath, working roll and prop up The anticipated relation calculated between inclination and both sides roll-force difference such as support roll crown, strip running deviation value etc., and be fitted with multinomial, According to each term coefficient in observed dip and both sides roll-force difference on-line amending multinomial model during online use.Roll according to both sides Power difference processed, calculates corresponding tilting value, and using this tilting value as setting the amplitude limit value of inclination, its principle is as shown in figure 14.

Embodiment: according to the inlet thickness of actual band steel, exit thickness, drafts, resistance of deformation, strip width, outlet Set tension force, inlet set tension force, working roll and support roller roller footpath, working roll and support roll crown, strip running deviation value etc. anticipated Calculate the relation between inclination and both sides roll-force difference, with roll-force difference as independent variable, to tilt as dependent variable, use multinomial Simulate the functional relation of roll-force difference and inclination

According to actually detected both sides roll-force difference DELTA E during application_DO, the tilting value of roll is calculated by above formula, and Using it as the amplitude limit setting tilting value.When band steel occurs that portion's crackle, raw material plate shape problem cause both sides Tension Difference or plate shape system System detecting element precise decreasing, causes tilt adjustments amount to be continuously increased, by effective amplitude limit, can limit tilt adjustments amount width Value, it is to avoid tilting value is excessive and causes the generation of broken belt accident, and its principle is shown in Figure 13.

For eliminating the error of theoretical calculation model, can be according to the actual roll-force of detection is poor and actual tilt data are to [Δ F_act(k), tilt_act(k)] (k=1,2,3 ...), by least square method, incline according to error sum of squares minimum principle dynamic corrections Each term coefficient in tiltedly-roll-force difference multinomial, revised multinomial isRevise next time Front inclination amplitude limit value, can be according to revised polynomial computation.

Claims (3)

1. cold-rolling mill Roll-gap position pressure double-closed-loop control method, it is characterised in that after milling train additional tilt, bilateral is asymmetric rolls Power processed calculates；Derived according to Ka Shi energy theorem be suitable to that Asymmetric Rolling calculates support roller, working roll simply supported beam form Elastic bending influence function；Rolls' deformation theory effectively integrates with metal theory of cross-flow；To conventional roll system Deformation Theory In Coordinate deformation equation be effectively improved；The error calculated in view of theory, is carried out theoretical calculation model in real time Online adaptive correction；

After rolling mill additional tilt, contact in rolling is the most no longer traditional left and right centered by rolling centerline Symmetry, but asymmetrical, when carrying out asymmetric calculating, the roll-force of the left and right sides is unknown number；When calculating, unit is drawn Being divided into and arranging from left to right along roller whole body, support bending influence function of deriving is:

In formula, E_bFor supporting roller elastic modelling quantity；I_b1For supporting the bending resistant section coefficient in roll shaft footpath；I_b2For supporting the bending resistance of the roller body of roll Section factor；Φ is shearing deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_b1Transversal for support at roll shaft footpath Area；A_b2For supporting the cross-sectional area of the roller body of roll；l_bFor supporting roller two ends strong point spacing；l_OFor the axle head strong point to body of roll end The distance in face；x_iFor the virtual force application point distance away from the axle head strong point；x_jFor the unit point of force application distance away from the axle head strong point； l_DFor the axle head strong point distance away from opposite side end surface of roll；

Working roller bending influence function is:

In formula, E_wFor working roll elastic modelling quantity；I_w1Bending resistant section coefficient for the working roll diameter of axle；I_w2Bending resistance for the working roll body of roll Section factor；Φ is shearing deformation energy coefficient, Φ=10//9；G is the coefficient of rigidity；A_w1Transversal at the working roll diameter of axle Area；A_w2Cross-sectional area for the working roll body of roll；l_wFor working roll two ends support distances；x_iSupport away from axle head for virtual force application point The distance of point；x_jFor the unit point of force application distance away from the axle head strong point；l_OWThe fore side strong point is away from fore side end surface of roll Distance；l_DWFor the fore side strong point distance away from transmission side end surface of roll；

Coordinate deformation equation between working roll and support roller is:

In formula

For supporting the tilt vector of roller；

For flattening vector between roller,

Constant vector, i.e. the flattening amount of roll surface center；

It is to support roll crown vector；

It it is work roll crown vector；

NB is for supporting roller cutting unit quantity；For working roller bending amount of deflection vector.

Cold-rolling mill Roll-gap position pressure double-closed-loop control method the most according to claim 1, it is characterised in that additional tilt Calculated by rolls' deformation and deformed area metal lateral flow afterwards, finally can obtain both sides roll-force and difference is:

Transmission side roll-force:

Fore side roll-force is:

Both sides roll-force difference is: Δ F=F_D-F_O

L in formula_bFor supporting roller two ends strong point spacing；NB is for supporting roller cutting unit quantity；Q (i) is that i-th cutting unit is made Load；X (i) is the i-th cutting unit distance away from the fore side strong point；With roll-force difference as independent variable, to tilt Value is dependent variable, and the functional relation going out roll-force difference and tilting value with fitting of a polynomial is

In formula: tilt is tilting value；a₀I () is the coefficient of polynomial fitting i & lt item；N is the high reps of polynomial fitting.

Cold-rolling mill Roll-gap position pressure double-closed-loop control method the most according to claim 1, it is characterised in that for eliminating reason The error of opinion computation model, can be according to the actual roll-force of detection is poor and actual tilt data are to [Δ F_act(k), tilt_act(k)] K=1,2,3 ..., by least square method, according in error sum of squares minimum principle dynamic corrections inclination-roll-force difference multinomial Each term coefficient, after correction, the functional relation of roll-force difference and tilting value is

In formula: tilt is tilting value；a₁I () is the coefficient of revised polynomial fitting i & lt item；N be polynomial fitting High reps；Δ F is the difference of transmission side rolling and fore side roll-force.