CN102728624B - Method for setting load distribution of finish rolling band steel - Google Patents

Abstract

The invention relates to a method for setting the load distribution of finish rolling band steel, belonging to the field of production control of finish rolling band steel, and solving the problems that the rolling force proportion of stands is large in fluctuation when the rolling condition is changed under the pressed-down distribution mode, and the thickness distribution of the stands is large in fluctuation when the rolling force model is large in deviation under the rolling force distribution mode. According to the invention, the pressed-down distribution mode and the rolling force distribution mode are combined; the rolling force distribution result is restricted by the distribution range determined by pressed-down distribution; the requirement that the rolling force of stands is decreased according to proportion is met while the thickness distribution fluctuation of the stands is reduced; the results of two distribution modes are gradually consistent, and the iterations are gradually reduced by learning a pressed-down rate learning table when the deviation of the rolling force model is small; and the fluctuation range of the pressed-down rate of the stands is effectively controlled through the pressed-down distribution amplitude limit when the deviation of the rolling force model is large. By adopting the method, the advantages of the two modes are combined and the shortages are overcome, so that the rolling schedule setting precision and the rolling stability of the finish rolling band steel are effectively improved.

Description

A kind of fine-rolling strip steel sharing of load establishing method

Technical field

The present invention relates to fine-rolling strip steel production control field, relate in particular to a kind of fine-rolling strip steel sharing of load establishing method.

Background technology

In hot strip rolling unit, the formulation of rolling procedure plays an important role for improving the quality of products, its rationally whether, the stability of the height to product quality, the difficulty or ease of rolling equipment adjustment, unit maintainability and production process all has a direct impact.Rolling procedure generally comprises reduction system, speed system, tension schedule, temperature schedule etc.Wherein reduction system refers to the distribution system of each frame (or each passage) drafts, is conventionally referred to as in the industry sharing of load or thickness and distributes.

Sharing of load is the core of rolling procedure, and it directly has influence on the product quality such as plate shape, thickness of slab precision, and sharing of load also has material impact to the item such as stability and the operating rate index of required power, roller consumption, production process.The load distribution method of hot strip rolling unit from before experience schedule method, also experienced several stages such as energy consumption method, dynamic load distribution method, sharing of load Y-factor method Y, artificial intelligence approach.

At present, the method that modernization hot continuous rolling unit sharing of load more generally adopts is sharing of load Y-factor method Y, and it is by drafts and the rolled piece thickness of certain given each frame of sharing of load coefficient.Sharing of load coefficient can Reference cultivars and specification determine, also can adopt the method for offline optimization to determine.Sharing of load Y-factor method Y mainly comprises the types such as drafts, reduction ratio, roll-force and power.Wherein roll-force, power type need iterative computation, are all classified as roll-force allocation model for convenience of analyzing; Drafts, reduction ratio type can directly obtain drafts and the thickness of each frame, are all classified as and depress allocation model.The pluses and minuses of these two kinds of patterns:

The advantage of depressing distribution is that each frame thickness distributes fluctuation little, and computational process is simple; When shortcoming is rolling situation change (as roll change), each frame roll-force ratio fluctuation greatly, can not be eliminated roll-force reversal of the natural order of things, can not ensure production board shape.

The advantage that roll-force is distributed is each frame roll-force down ratio substantially constant, is conducive to production board shape and rolling stability; Shortcoming is that each frame thickness distributes fluctuation greatly, computational process complexity in the time that rolling force model deviation is large or process conditions change greatly.

At present, about also there are some disclosed patent documents in the establishing method of fine-rolling strip steel sharing of load, example, name is called the method for recording in " a kind of method of dynamically setting tandem rolling schedule of hot rolling strip " file (CN101733289A): will depress distribution coefficient as optimizing starting point, calculate now and load accordingly, and compare with target load distribution coefficient, adjust each reduction in pass according to difference; Through repeatedly comparing and adjustment, reach assumed load and distribute the object of approaching target load distribution.Its limitation is: 1. only adjust the distribution of two frames, between frame, harmony is bad at every turn, and regulated efficiency is lower; 2. only will depress distribution coefficient as optimizing starting point, still a kind of roll-force is distributed in essence, can not overcome the shortcoming that roll-force is distributed.

Summary of the invention

The object of the invention is the problem existing in order to solve prior art, propose a kind of fine-rolling strip steel sharing of load establishing method, thereby improve fine-rolling strip steel rolling procedure setting accuracy and rolling stability.

The object of the invention is to be achieved through the following technical solutions:

A kind of fine-rolling strip steel sharing of load establishing method, comprises the following steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table, by the allocation result of depressing of depressing allocation model and calculate finish rolling each frame, described in depress the reduction ratio distributing under allocation result finger pressure with the exit thickness of depressing distribution s2, each frame obtained above is depressed to the exit thickness of distribution as initial value, by the absolute draft amount of the each frame of roll-force allocation model iterative computation, until roll-force ratio meets target call, obtain the roll-force allocation result of each frame, described roll-force allocation result comprises the exit thickness that roll-force is distributed and the reduction ratio of roll-force distribution s3, the roll-force allocation result of described each frame is carried out to amplitude limiting processing, make it depress allocation result within the specific limits described in departing from, thereby the final reduction ratio that makes each frame is distributing in definite certain limit by depressing, just get roll-force allocation result as long as do not go beyond the scope, go beyond the scope and be limited in boundary; S4, band steel head threading completes and collects after actual roll-force, judges the deviation of each frame rolling force setup value and actual roll-force; S5, judge that each frame deviate is whether all in θ, θ is Deviation Control parameter, get 5%-10%, if do not meet all conditions in θ of each frame rolling force deviation, do not carry out reduction ratio study, now, in described reduction ratio learning table, other data of equivalent layer are not upgraded, and flow process enters end step; S6, if each frame deviate, in θ, is learnt the final reduction ratio of this setting, is updated to reduction ratio study form equivalent layer by learning value other, and through learning process, this layer of other roll-force can progressively reach the requirement of target roll-force ratio; S7, sharing of load is set and is finished.

Described step S1 comprises the following steps:

S11, gets table, gets reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table, and both are added, and obtain the reduction ratio allocation table scale value of each frame

S12, carries out relativization calculating, obtains the allocation result of depressing of each frame: must equal the overall reduction of finish rolling according to the drafts sum of the each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{xrk}=\frac{-\mathrm{kb}\·{\mathrm{xs}}_1+\sqrt{{(\mathrm{kb}\·{\mathrm{xs}}_1)}^2+4\·\mathrm{ka}\·{\mathrm{xs}}_2\·1\mathrm{nx}}}{2\·\mathrm{ka}\·{\mathrm{xs}}_2}$ Formula 1

Wherein, h _nfor finish to gauge thickness, h ₀for workpiece thickness;

n is finishing pass number, and i is shelf number;

kb=0.906501，ka=0.959597。

Reduction ratio after the reduction ratio allocation table scale value relativization of each frame is the reduction ratio of depressing distribution

${\mathrm{\ϵ}}_{\mathrm{red}}^i=\mathrm{xrk}\·{\mathrm{\ϵ}}_0^i$ Formula 2

According to the reduction ratio of depressing distribution obtain above n-1 the frame exit thickness of depressing distribution separately $h_{\mathrm{red}}^i:$

$\begin{array}{cc}{h}_{\mathrm{red}}^i=h_0\·\underset{i=1}{\overset{i}{\mathrm{\Π}}}(1-{\mathrm{\ϵ}}_{\mathrm{red}}^i)& i=\mathrm{1,2},...,n-1.\end{array}$ Formula 3

The reduction ratio allocation table scale value of each frame in described step S11 the process of obtaining is: according to the large class of the current steel grade with steel, level of thickness and width class information, do not get the reduction ratio of the each frame of finish rolling from depressing distribution coefficient table equivalent layer wherein i is shelf number; According to current tapping mark, level of thickness and width class information with steel, do not get the reduction ratio learning value of the each frame of finish rolling from reduction ratio learning table equivalent layer wherein i is shelf number, and described reduction ratio learning table is the tables of data being obtained by program automatic learning, and in table, data are constantly updated by study, and initial value gets 0; The reduction ratio allocation table scale value of described each frame for:

${\mathrm{\ϵ}}_0^i={\mathrm{\ϵ}}_{\mathrm{ori}}^i+{\mathrm{\ϵ}}_{\mathrm{lrn}}^i.$

Described step S2 comprises the following steps:

S21, according to the large class of the current steel grade with steel and level of thickness information, does not get the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i;

S22, operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i: the operation modifying factor ζ that collects each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1ifor:

${\mathrm{\α}}_{1i}={\mathrm{\α}}_{0i}\·(1+{\mathrm{\ζ}}_i)$ Formula 4

Wherein, ζ _ispan is-20%～20%;

Be normalized, the roll-force distribution coefficient α i after normalization is:

S23, roll-force and the derivative of roll-force to absolute draft amount of calculating each frame under current thickness distributes: calculate roll-force if: i is shelf number, and j is iterations, be the roll-force of the j time iteration of i frame, the exit thickness of the j time iteration of i frame, the exit thickness of the j time iteration of i-1 frame; The front i frame of iterative computation exit thickness initial value for the first time the exit thickness of depressing distribution obtaining for above-mentioned formula (3) in the situation that other technological parameter is fixing, roll-force is the function of inlet thickness and exit thickness, exists:

Calculate the derivative of roll-force to absolute draft amount

Wherein, for the roll-force after the positive disturbance of absolute draft amount; for the roll-force after the negative disturbance of absolute draft amount, it is the absolute draft amount of the j time iteration of i frame; There is following functional relation: $p_{i1}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\Δh}}_{i1}^j),p_{i2}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\Δh}}_{i2}^j),$ Wherein, for the absolute draft amount after positive disturbance, for the absolute draft amount after negative disturbance,

${\mathrm{\Δh}}_{i1}^j=(1+\frac{\mathrm{dt}}{2})\·{\mathrm{\Δh}}_i^j,{\mathrm{\Δh}}_{i2}^j=(1-\frac{\mathrm{dt}}{2})\·{\mathrm{\Δh}}_i^j,$ Wherein dt is disturbance, gets 1%;

S24, the result of calculation to absolute draft amount derivative according to the roll-force of above-mentioned each frame and roll-force, calculate the correction value of each frame absolute draft amount:

Rolling distributes iterative computation formula:

$\mathrm{\δ}\left({\mathrm{\Δh}}_i^j\right)=({\mathrm{\Σp}}_i^j\·\frac{{\mathrm{\α}}_i}{\mathrm{\Σ}{\mathrm{\α}}_i}-p_i^j)/\frac{{\∂p}_i^j}{\∂\left(\mathrm{\Δ}{h}_i^j\right)}-\mathrm{\Σ}((\mathrm{\Σ}{p}_i^j\·\frac{{\mathrm{\α}}_i}{\mathrm{\Σ}{\mathrm{\α}}_i}-p_i^j)/\frac{{\∂p}_i^j}{\∂\left({\mathrm{\Δh}}_i^j\right)})\·\frac{{\mathrm{\Δh}}_i^j}{\mathrm{\Σ\Δ}{h}_i^j}$ Formula 5

Wherein: it is the correction value of the absolute draft amount of the j time iteration of i frame;

S25, the correction value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame absolute draft amount, the absolute draft amount that this renewal value is the j+1 time iteration computing formula is:

${\mathrm{\Δh}}_i^{j+1}={\mathrm{\Δh}}_i^j+{\mathrm{damp}}_j\·\mathrm{\δ}\left({\mathrm{\Δh}}_i^j\right)$ Formula 6

Wherein damp _jfor damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy gets 1.0, β and gets 0.6;

S26, the renewal value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame exit thickness, the exit thickness that this renewal value is the j+1 time iteration

and workpiece thickness h ₀for known quantity, according to only first frame all the other are all the inputs that previous frame is output as a rear frame, calculate last frame from first frame, obtain the renewal value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force use, wherein refer to the exit thickness of the j+1 time iteration of i-1 frame;

S27, judges whether roll-force ratio meets the condition of convergence:

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\Σ}{p}_i^j}-\frac{{\mathrm{\α}}_i}{\mathrm{\Σ}{\mathrm{\α}}_i}|\≤\frac{\mathrm{\τ}}{{\mathrm{\Σ\α}}_i}$ Formula 7

Wherein, τ is positive number, gets 0.01; When formula (7) is set up, or iterations j while exceeding set point number finishing iteration calculate, otherwise continue to carry out described S23 step, Simultaneous Iteration number of times j cumulative 1;

S28, if roll-force ratio meets the condition of convergence in above-mentioned steps S27, judges whether the roll-force of certain frame exceedes the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

As the roll-force p of i frame _iexceed the maximum rolling force that equipment allows time, enter S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

time, carry out ${\mathrm{\α}}_i^{\′}={\mathrm{\α}}_i\·\frac{1}{(p_i/p_i^{\max }-1)\·0.8+1}$ Formula 8

Wherein, α _ifor the roll-force distribution coefficient α after described normalization _i, α _i' be the roll-force distribution coefficient transfiniting after adjusting;

α _i' get back to step S23 after obtaining, with after adjusting roll-force distribution coefficient α _i' re-starting absolute draft amount iterative computation, Simultaneous Iteration number of times j assignment 0, gets the absolute draft amount of the last iteration of step S27 and exit thickness as the initial value of absolute draft amount and exit thickness;

If there is no roll-force transfinites, and directly carries out next step;

S29, absolute draft amount iterative computation and the inspection of transfiniting finish, and obtain roll-force allocation result, get the exit thickness of the last iteration of step S27 the exit thickness distributing as each frame roll-force obtain the reduction ratio that each frame roll-force is distributed simultaneously for: in formula be the exit thickness that i frame roll-force is distributed, it is the exit thickness that i-1 frame roll-force is distributed.

Described step 3 comprises the following steps:

S31, carries out amplitude limiting processing to roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{xf}}^i=\left\{\begin{array}{cc}{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i& {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.$ Formula 9

Wherein for reduction ratio after amplitude limiting processing, η is amplitude limit control parameter, gets 5%～10%;

S32, to reduction ratio after amplitude limiting processing carry out relativization calculating:

If reduction ratio is modified in above-mentioned steps S31, calculating finish to gauge thickness h ' _nwith target finish to gauge thickness h _nbetween can there is certain deviation, also to carry out taking turns relativization calculate:

Calculating finish to gauge thickness h ' _n:

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{1n\frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after finish to gauge relativization be called final distribution reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get as the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously

$\begin{array}{cc}{h}_{\mathrm{xdh}}^i:h_{\mathrm{xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i)& i=\mathrm{1,2},...,n-1,\end{array}$

According to final distribution reduction ratio with final outlet thickness carry out the rolling procedure of fine-rolling strip steel and set calculating.

In described step S6, the process that final reduction ratio is learnt is:

Ask that each frame is final distributes and reduction ratio poor of depressing distribution

${\mathrm{\&epsiv;}}_{\mathrm{dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i}{\mathrm{xrk}}$ Formula 10

Wherein, for the final reduction ratio distributing, for depressing the reduction ratio of distribution, xrk is the scale factor of the reduction ratio in formula (1);

Obtain rolling procedure and set the learning value that rear reduction ratio learning table equivalent layer does not upgrade

${\mathrm{\&epsiv;}}_{\mathrm{lrn}}^{i,\mathrm{new}}={\mathrm{\&epsiv;}}_{\mathrm{lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{dif}}^i\&CenterDot;(1-\mathrm{\&mu;})$ Formula 11

Wherein, μ is smoothing factor, gets 0.7; for rolling procedure is set the front not original learning value of reduction ratio learning table equivalent layer.

Beneficial effect of the present invention:

Fine-rolling strip steel sharing of load establishing method of the present invention, realize the aggregative model of " depressing distribution+roll-force distributes ", distribute definite distribution to retrain roll-force allocation result with depressing, combine the advantage of two kinds of patterns, overcome both shortcomings separately simultaneously, reduce in each frame thickness distributes fluctuation to meet the requirement that each frame roll-force declines in proportion as far as possible, reach the object that improves fine-rolling strip steel rolling procedure setting accuracy and rolling stability.

For further illustrating above-mentioned purpose of the present invention, design feature and effect, below with reference to accompanying drawing, the present invention is described in detail.

Brief description of the drawings

Fig. 1 is fine-rolling strip steel sharing of load establishing method flow chart of the present invention;

Fig. 2 is the particular flow sheet that calculates each frame absolute draft amount in Fig. 1 flow process;

Fig. 3 is that roll-force is distributed and the final reduction ratio being caused by rolling force model deviation that distributes fluctuates.

Detailed description of the invention

Accompanying drawing below in conjunction with embodiment is elaborated to the specific embodiment of the present invention.

Referring to Fig. 1, Fig. 1 is fine-rolling strip steel sharing of load establishing method flow chart of the present invention, and fine-rolling strip steel sharing of load establishing method of the present invention, comprises the following steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table, by the allocation result of depressing of depressing allocation model and calculate finish rolling each frame, described in depress the reduction ratio distributing under allocation result finger pressure with the exit thickness of depressing distribution it comprises the following steps:

S11, gets table, gets reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table, and both are added, and obtain the reduction ratio allocation table scale value of each frame, and detailed process is:

According to the large class of the current steel grade with steel, level of thickness and width class information, do not get the reduction ratio of the each frame of finish rolling (example: F1～F7) from depressing distribution coefficient table equivalent layer (i is shelf number), described reduction ratio distribution coefficient table is to commonly use in the industry table, in table, data are determined before production, are generally the empirical data of considering capacity of equipment and product performance, can collect the excellent case in actual production;

According to current tapping mark, level of thickness and width class information with steel, do not get the reduction ratio learning value of the each frame of finish rolling (F1～F7) from reduction ratio learning table equivalent layer (i is shelf number), described reduction ratio learning table is the tables of data being obtained by program automatic learning, and in table, data are constantly updated by study, and initial value gets 0;

Both additions are obtained to the reduction ratio allocation table scale value of the each frame of finish rolling (i is shelf number).

The benefit of two-layer Table Design is can reflect by reduction ratio learning table the difference (a large class of steel grade comprises some tapping marks) of different tapping marks under the large class of same steel grade.

S12, carries out relativization calculating, obtains the allocation result of depressing of each frame, depresses the reduction ratio of distribution with the exit thickness of depressing distribution detailed process is:

In order to obtain the concrete reduction ratio data under various workpiece thickness and finish to gauge thickness, need to be to above-mentioned reduction ratio allocation table scale value carry out relativization calculating, must equal the overall reduction of finish rolling according to the drafts sum of the each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{xrk}=\frac{-\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1+\sqrt{{(\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1)}^2+4\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2\&CenterDot;1\mathrm{nx}}}{2\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2}---\left(1\right)$

Wherein, h _nfor finish to gauge thickness, h ₀for workpiece thickness (h _nwith h ₀that finish rolling rolling procedure is set the front known value of calculating);

n is finishing pass number, and i is shelf number;

kb=0.906501，ka=0.959597。

Reduction ratio (depressing the reduction ratio of distribution) after the reduction ratio allocation table scale value relativization of each frame for:

${\mathrm{\&epsiv;}}_{\mathrm{red}}^i=\mathrm{xrk}\&CenterDot;{\mathrm{\&epsiv;}}_0^i---\left(2\right)$

According to the reduction ratio of depressing distribution can obtain above n-1 the frame exit thickness of depressing distribution separately $h_{\mathrm{red}}^i:$

$\begin{array}{cc}{h}_{\mathrm{red}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i)& i=\mathrm{1,2},...,n-1.\end{array}---\left(3\right)$

Formula (3) is to obtain according to the relation between each frame inlet thickness, exit thickness and reduction ratio, is the fundamental formular using in the industry.

S2, depress the exit thickness of distribution with the each frame obtaining in above-mentioned S12 for initial value, by the absolute draft amount of the each frame of roll-force allocation model iterative computation, until roll-force ratio meets target call, obtain the roll-force allocation result of each frame, described roll-force allocation result comprises the exit thickness that roll-force is distributed and the reduction ratio of roll-force distribution idiographic flow is referring to Fig. 2, comprising absolute draft amount iterative process (S21-S27) and the checking process that transfinites (S28):

S21, according to the large class of the current steel grade with steel and level of thickness information, does not get the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i.

In above-mentioned distribution coefficient table, data are determined before production, mainly consider the factor such as belt steel rolling stability and plate shape index, can on empirical data basis, constantly collect outstanding rolling case and be optimized.Reduction ratio distribution coefficient table in above-mentioned steps S1 is necessary by depressing allocation model, roll-force distribution coefficient table in this step by roll-force allocation model necessary, also be known in the industry tables of data, both data are independently, be all technique empirical data, the present invention has realized the comprehensive of these two kinds of distribution.

S22, operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i.

Collect the operation modifying factor ζ of each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1ifor:

α _1i=α _0i·(1+ζ _i) (4)

Wherein, ζ _ispan is-20%～20%.

Be normalized the roll-force distribution coefficient α after normalization _ifor:

The normalization here refers to α maximum in institute's organic frame (routine F1～F7 is totally 7 frames) _1ivalue is classified as 1, and all the other projects are carried out respective handling.

S23, roll-force and the derivative of roll-force to absolute draft amount of calculating each frame under current thickness distributes:

(1) calculate roll-force

The variable declaration arranging in formula below: i is shelf number, and j is iterations, be the roll-force of the j time iteration of i frame, the exit thickness of the j time iteration of i frame, the exit thickness of the j time iteration of i-1 frame.

The front i frame of iterative computation exit thickness initial value for the first time the exit thickness of depressing distribution obtaining for above-mentioned formula (3)

In the situation that other technological parameter is fixing, roll-force is the function of inlet thickness and exit thickness, exists: (this functional relation, for known in the industry, does not explain at this), the size of roll-force is subject to the impact of rolling force model precision; Be noted that, the variation that each frame thickness distributes can cause the variation of each frame temperature, and calculating before each frame roll-force, (temperature and roll-force interact should to recalculate once the temperature of each frame, its interactive pass be in the industry known to, do not explain at this);

(2) calculate the derivative of roll-force to absolute draft amount

Computing formula is: the formula principle of this computational process be roll-force variable quantity divided by absolute draft amount variable quantity, i.e. the definition of derivative, is known computational process in the industry, is briefly described at this.

In formula, for the roll-force after the positive disturbance of absolute draft amount; for the roll-force after the negative disturbance of absolute draft amount, it is the absolute draft amount of the j time iteration of i frame;

There is following functional relation: $p_{i1}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\&Delta;h}}_{i1}^j),p_{i2}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\&Delta;h}}_{i2}^j)$

Wherein, for the absolute draft amount after positive disturbance, for the absolute draft amount after negative disturbance,

${\mathrm{\&Delta;h}}_{i1}^j=(1+\frac{\mathrm{dt}}{2})\&CenterDot;{\mathrm{\&Delta;h}}_i^j,{\mathrm{\&Delta;h}}_{i2}^j=(1-\frac{\mathrm{dt}}{2}){\&CenterDot;\mathrm{\&Delta;h}}_i^j$

Wherein dt is very little disturbance, desirable 1%.

S24, the result of calculation to absolute draft amount derivative according to the roll-force of above-mentioned each frame and roll-force, calculates the correction value of each frame absolute draft amount.

Rolling distributes the core formula of iterative computation:

$\mathrm{\&delta;}\left({\mathrm{\&Delta;h}}_i^j\right)=({\mathrm{\&Sigma;p}}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{{\&PartialD;p}_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}-\mathrm{\&Sigma;}((\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{{\&PartialD;p}_i^j}{\&PartialD;\left({\mathrm{\&Delta;h}}_i^j\right)})\&CenterDot;\frac{{\mathrm{\&Delta;h}}_i^j}{\mathrm{\&Sigma;\&Delta;}{h}_i^j}---\left(5\right)$

Wherein: it is the correction value of the absolute draft amount of the j time iteration of i frame.

Above-mentioned formula (5) is the result deriving according to a kind of Newton of improvement method, select the absolute draft amount of each passage as variable, the nonlinear equation of n dimension is become to n nonlinear equation of one dimension, and each nonlinear equation is solved by Newton iterative method.It has used for reference Newton-Raphson method the feature of good convergence, can avoid again asking Jacobian matrix and inverse matrix thereof simultaneously, and algorithm is effectively simplified, and computational speed is very fast, and precision is higher.

S25, the correction value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame absolute draft amount, namely the absolute draft amount of the j+1 time iteration

Computing formula is:

$\mathrm{\&Delta;}{h}_i^{j+1}=\mathrm{\&Delta;}{h}_i^j+{\mathrm{damp}}_j\&CenterDot;\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)---\left(6\right)$

Wherein damp _jfor damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy is desirable 1.0, β desirable 0.6.

When each iteration, only absorb the certain percentage of absolute draft amount correction value, along with iterations increases, percent absorption also increases.

S26, the renewal value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame exit thickness, namely the exit thickness of the j+1 time iteration

and workpiece thickness h ₀for known quantity, according to only first frame all the other are all the inputs that previous frame is output as a rear frame, calculate last frame from first frame, obtain the renewal value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force use, wherein refer to the exit thickness of the j+1 time iteration of i-1 frame;

Routine to F1 frame, i=1, no matter which time iteration, has constant;

To F2 frame, i=2, f1 output is inputted as F2;

To F3 frame, i=3, successively recursion .....

So be summarized as only in the time of F1, all the other are all the inputs that previous frame is output as a rear frame.

S27, judges whether roll-force ratio meets the condition of convergence.

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\&Sigma;}{p}_i^j}-\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}|\&le;\frac{\mathrm{\&tau;}}{{\mathrm{\&Sigma;\&alpha;}}_i}---\left(7\right)$

Wherein, τ is very little positive number, desirable 0.01.

When formula (7) is set up, or iterations j while exceeding set point number finishing iteration calculate, otherwise continue to carry out S23 step, Simultaneous Iteration number of times j cumulative 1.For example, the set point number of iterations j is 6 times, and the span of j is 0≤j≤5, and the 1st time iterative computation j gets 0, when formula (7) set up or when iterations j exceedes 6 times iterative computation finish; If iterative computation does not also reach the condition of convergence for the first time, continue to carry out S23 step, Simultaneous Iteration number of times j cumulative 1.

After iterative computation finishes, can obtain the absolute draft amount of each frame and exit thickness

S28, if roll-force ratio meets the condition of convergence in above-mentioned steps S27, judges whether the roll-force of certain frame exceedes the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

When the roll-force pi of i frame exceedes the maximum rolling force that equipment allows time, enter S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

time, carry out ${\mathrm{\&alpha;}}_i^{\&prime;}={\mathrm{\&alpha;}}_i\&CenterDot;\frac{1}{(p_i/p_i^{\max }-1)\&CenterDot;0.8+1}---\left(8\right)$

Wherein, α _ifor the roll-force distribution coefficient α after described normalization _i, α _i' be the roll-force distribution coefficient transfiniting after adjusting;

α _i' get back to step S23 after obtaining, with after adjusting roll-force distribution coefficient α _i' re-start absolute draft amount iterative computation.Simultaneous Iteration number of times j assignment 0, gets the absolute draft amount of the last iteration of S27 and exit thickness as the initial value of absolute draft amount and exit thickness;

If there is no roll-force transfinites, and directly carries out next step S29.

S29, absolute draft amount iterative computation and the inspection of transfiniting finish, and obtain roll-force allocation result, get the exit thickness of the last iteration of step S27 the exit thickness distributing as each frame roll-force obtain the reduction ratio that each frame roll-force is distributed simultaneously for: in formula be the exit thickness that i frame roll-force is distributed, it is the exit thickness that i-1 frame roll-force is distributed.

In S2 step, can reduce iterations to depress the initial value that allocation result distributes as roll-force, roll-force is distributed can control the ratio that each frame roll-force declines, and then improves rolling stability and the plate shape index of fine-rolling strip steel.

S3, the roll-force allocation result of each frame in above-mentioned steps S2 is carried out to amplitude limiting processing, make it depart from above-mentioned steps S1 and depress allocation result within the specific limits, thereby the final reduction ratio that makes each frame is distributing in definite certain limit by depressing, just get roll-force allocation result as long as do not go beyond the scope, go beyond the scope and be limited in boundary.

S31, carries out amplitude limiting processing to roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{xf}}^i=\left\{\begin{array}{cc}{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i& {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.---\left(9\right)$

Wherein for reduction ratio after amplitude limiting processing, η is amplitude limit control parameter, gets 5%～10%;

S32, to reduction ratio after amplitude limiting processing carry out relativization calculating:

If reduction ratio is modified in above-mentioned steps S31, calculate finish to gauge thickness with target finish to gauge thickness h _nbetween can there is certain deviation, also to carry out taking turns relativization calculate:

Calculate finish to gauge thickness

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{1n\frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after finish to gauge relativization be called final distribution reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get as the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously

$\begin{array}{cc}{h}_{\mathrm{xdh}}^i:h_{\mathrm{xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i)& i=\mathrm{1,2},...,n-1,\end{array}$

According to final distribution reduction ratio with final outlet thickness carry out the rolling procedure of fine-rolling strip steel and set calculating.

S4, band steel head threading completes and collects after actual roll-force, judge the deviation of rolling force setup value (rolling force model of this rolling force setup value during by rolling schedule calculation calculates, and is publicly-owned knowledge, with calculate roll-force in above-mentioned S23 the same) and actual roll-force.

S5, judge that each frame deviate is whether all in θ, θ is Deviation Control parameter, desirable 5%-10%, if do not meet all conditions in θ of each frame rolling force deviation, do not carry out reduction ratio study, now, in reduction ratio learning table (following table 2), other data of equivalent layer are not upgraded, and flow process enters end step.

S6, if each frame deviate, in θ, is learnt the final reduction ratio of this setting, it is other that learning value is updated to reduction ratio study form equivalent layer, for:

${\mathrm{\&epsiv;}}_{\mathrm{dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i}{\mathrm{xrk}}---\left(10\right)$

Wherein be the poor of the final distribution of i frame and the reduction ratio of depressing distribution, xrk is the scale factor of the reduction ratio in formula (1).

Obtain rolling procedure and set the learning value that rear reduction ratio learning table equivalent layer does not upgrade for:

${\mathrm{\&epsiv;}}_{\mathrm{lrn}}^{i,\mathrm{new}}={\mathrm{\&epsiv;}}_{\mathrm{lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{dif}}^i\&CenterDot;(1-\mathrm{\&mu;})---\left(11\right)$

Wherein, μ is smoothing factor, desirable 0.7; for the not original learning value of reduction ratio learning table (following table 2) equivalent layer (before rolling procedure setting).

S7, calculates and finishes, and is inciting somebody to action be updated to reduction ratio learning table (form 2) equivalent layer not after, due to study effect, this layer of other roll-force can progressively reach the requirement of target roll-force ratio.

Below for embodiment explanation.

Embodiment 1:

Stress whole calculation process.Final result is to distribute through depressing the roll-force of distributing after amplitude limit, overcome in the time that alone roll-force is distributed large drawback (this kind of situation generation when large and process conditions change greatly in rolling force model error of fluctuating of frame reduction ratio, roll-force is distributed the requirement that will reach by adjusting reduction ratio each frame roll-force ratio, so reduction ratio fluctuation is large; Depress a point reduction ratio for the each frame of timing and almost remain unchanged and use), be conducive to raising fine-rolling strip steel rolling procedure setting accuracy and rolling stability.

Certain band steel information is as follows: workpiece thickness 40.69mm, and finish to gauge thickness 2.0mm, strip width 1233.6mm, finish rolling inlet temperature FET is 1000 DEG C, finish to gauge target temperature FDT is 880 DEG C.Finish rolling rolling pass is counted n=7.The other information of layer: the large class of steel grade is 11002, and level of thickness is 6, and width grade is 3, and tapping mark is GV4924E1.Get other data of equivalent layer from 3 forms as follows:

Table 1: reduction ratio distribution coefficient table

Table 2: reduction ratio learning table:

Table 3: roll-force distribution coefficient table

Above-mentioned steps S1: according to depressing distribution coefficient table, by depress allocation model calculate each frame thickness distribute by depress distribution coefficient table and reduction ratio learning table data be added, obtain following table 4: depress allocation table scale value

Be that 40.69mm and finish to gauge thickness are the concrete reduction ratio data under 2.0mm in order to obtain at workpiece thickness, it is right to need carry out relativization calculating.

According to formula (1), calculate the scale factor xrk of reduction ratio

$x=\frac{h_0}{h_7}=20.346,{\mathrm{xs}}_1=\underset{i=1}{\overset{7}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_0^i=2.36,{\mathrm{xs}}_2=\underset{i=1}{\overset{7}{\mathrm{\&Sigma;}}}{\left({\mathrm{\&epsiv;}}_0^i\right)}^2=0.9768$

$\mathrm{xrk}=\frac{-\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1+\sqrt{{(\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1)}^2+4\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2\&CenterDot;1\mathrm{nx}}}{2\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2}=0.984$

Obtain following table 5: the reduction ratio after each frame relativization and exit thickness data

Above-mentioned steps S2: to depress allocation result as initial value, obtain the roll-force allocation result of each frame.

First carry out absolute draft amount iterative process (S23～S27), obtain following table 6: the roll-force ratio of each iteration changes

Iterations	F1	F2	F3	F4	F5	F6	F7
								1	1	0.90617	0.85469	0.75396	0.55891	0.45492	0.31708
2	1	0.93258	0.86329	0.75989	0.5853	0.45067	0.33803
								3	1	0.95718	0.87182	0.76879	0.60745	0.44919	0.35346
4	1	0.97063	0.87697	0.7759	0.61707	0.44984	0.35862
								Target	1	0.98	0.88	0.78	0.62	0.45	0.36

Carry out the checking process (S28) that transfinites, each frame roll-force does not all exceed the maximum rolling force of its permission, therefore do not need to come back to S23 again.

Following table 7: reduction ratio and exit thickness that roll-force is distributed

Above-mentioned steps S3: roll-force allocation result is done to amplitude limiting processing, it is departed from and depress allocation result within the specific limits.

1, roll-force allocation result is done to amplitude limiting processing.Get η=6%, find institute organic frame roll-force allocation result all depress allocation result ± 6% in, therefore that thickness distributes is constant.

2, because data do not change in above-mentioned 1, therefore calculate without relativization.

Above-mentioned steps S4-S6: band steel head threading completes and collects after actual roll-force, judges the deviation of rolling force setup value and measured value, if each frame deviation all in θ, is learnt the reduction ratio of this setting.

Band steel head threading completes and collects after actual roll-force, judges the deviation of rolling force setup value and measured value, finds that each frame deviation all (gets θ=5%) in 5%, obtains following table 8: carry out reduction ratio study

Will be updated to equivalent layer in reduction ratio learning table other.Due to the effect of study, this layer does not depress distribution technique table and roll-force distribution technique table data can be more and more identical, depresses roll-force corresponding to allocation result and can more approach the requirement that roll-force is distributed target proportion.

This situation is ideal, and not only roll-force distribution reaches target proportion requirement, and departs from and depress allocation result within the specific limits, and without amplitude limiting processing, get the result that roll-force is distributed final distribution.

Suppose below same embodiment, input data are constant, and only rolling force model deviation is increased to following table 9:

Identical computational process is carried out one time.

Step S1 result is constant.Because the deviation of rolling force model is different, step S2 result of calculation difference.Iteration just meets end condition 5 times, and the roll-force ratio of each iteration changes as following table 10:

Iterations	F1	F2	F3	F4	F5	F6	F7
								1	1	0.86302	0.7326	0.64625	0.5323	0.47658	0.27178
2	1	0.89134	0.77522	0.69524	0.57158	0.46032	0.3122
								3	1	0.92722	0.82495	0.74563	0.60658	0.45129	0.34476
4	1	0.95498	0.85953	0.77391	0.62107	0.45078	0.35705
								5	1	0.97054	0.87596	0.78306	0.62332	0.45095	0.3599
Target	1	0.98	0.88	0.78	0.62	0.45	0.36

The reduction ratio that roll-force is distributed and exit thickness are as following table 11:

Above-mentioned steps S3, carries out amplitude limiting processing to roll-force allocation result, find F1, F3, F4, F6, F7 frame all exceed depress allocation result ± 6%(η=6%), amplitude limiting processing result is as following table 12:

After amplitude limiting processing, will do relativization calculates.Carry out and meet termination condition 1 time, xc=0.0079, the reduction ratio of final output and exit thickness are as following table 13:

Finally, the roll-force distribution of front latter two situation is done and is contrasted as following table 14 with the final reduction ratio fluctuation distributing:

Fig. 3 is that roll-force is distributed and the final reduction ratio being caused by rolling force model deviation that distributes fluctuates.Visible, if directly roll-force is distributed as final distribution, the latter is very large with respect to the former fluctuation so, F6, F7 nearly 15%, all constant only because rolling force model deviation just can cause so large fluctuation in other condition, all unfavorable to rolling procedure setting calculating and rolling stability; But the final distribution after amplitude limiting processing, the latter is also little with respect to the former fluctuation, and basic controlling is in 5%.Here it is, and this patent proposes the reason of distributing definite distribution to retrain roll-force allocation result with depressing, conventional art or be simple " depressing distribution ", simple " roll-force distribution ", this patent has been realized " depressing distribution+roll-force distributes ", its advantage is: in the time that rolling force model deviation is little, make two kinds of allocation model results more and more consistent by the study of form 2, iterations is fewer and feweri; When large or process conditions change greatly in rolling force model deviation, effectively control the fluctuation range of each frame reduction ratio by depressing the amplitude limit interaction energy of distribution, be conducive to improve finish rolling code and set computational accuracy and rolling stability.

Embodiment 2:

Chassis equipment limit check.

Band steel information: workpiece thickness 42.308mm, finish to gauge thickness 4.597mm, road number of times n=7.The roll-force distribution coefficient requiring in step S2 is as following table 15:

SGF	THICK	α1	α2	α3	α4	α5	α6	α7
									11002	16	1	0.98	0.88	0.78	0.62	0.45	0.36

The roll-force inspection of transfiniting

For ease of relatively, calculate respectively two kinds of sharing of loads in situation: a kind of is that the maximum rolling force of F2 is not limited; A kind of is that the maximum rolling force of F2 is limited, and supposes comparison of computational results is as following table 16.

	F1	F2	F3	F4	F5	F6	F7
								α i	1	0.98	0.88	0.78	0.62	0.45	0.36
α iAfter amendment	1	0.94	0.88	0.78	0.62	0.45	0.36

Visible, in the time that maximum rolling force restriction is set, F2 frame roll-force meets

In sum, the present invention is directed to the deficiency of depressing distribution and roll-force distribution in prior art, a kind of new fine-rolling strip steel load distribution method is proposed, realize " depressing distribution+roll-force distributes ", distribute definite distribution to retrain roll-force allocation result with depressing, the advantage of comprehensive two kinds of patterns, overcome both shortcomings separately simultaneously, reduce in each frame thickness distributes fluctuation to meet the requirement that each frame roll-force declines in proportion as far as possible, reach the object that improves fine-rolling strip steel rolling procedure setting accuracy and rolling stability.Main innovate point explanation: from single step, be mainly step S2,3; In general, realize the combination of two kinds of allocation models, advantage: in the time that rolling force model deviation is little, make two kinds of allocation model results more and more consistent by the study of form 2, iterations is fewer and feweri; When large or process conditions change greatly in rolling force model deviation, effectively control the fluctuation range of each frame reduction ratio by depressing the amplitude limit interaction energy of distribution, be conducive to improve finish rolling code and set computational accuracy and rolling stability.

Those of ordinary skill in the art will be appreciated that, above embodiment is only for object of the present invention is described, and not as limitation of the invention, as long as in essential scope of the present invention, variation, modification to the above embodiment all will drop in the scope of claim of the present invention.

Claims (6)

1. a fine-rolling strip steel sharing of load establishing method, is characterized in that comprising the following steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table, by the allocation result of depressing of depressing allocation model and calculate finish rolling each frame, described in depress the reduction ratio distributing under allocation result finger pressure with the exit thickness of depressing distribution

S2, each frame obtained above is depressed to the exit thickness of distribution as initial value, by the absolute draft amount of the each frame of roll-force allocation model iterative computation, until roll-force ratio meets target call, obtain the roll-force allocation result of each frame, described roll-force allocation result comprises the exit thickness that roll-force is distributed and the reduction ratio of roll-force distribution

S3, the roll-force allocation result of described each frame is carried out to amplitude limiting processing, make it depress allocation result within the specific limits described in departing from, thereby the final reduction ratio that makes each frame is distributing in definite certain limit by depressing, just get roll-force allocation result as long as do not go beyond the scope, go beyond the scope and be limited in boundary;

S4, band steel head threading completes and collects after actual roll-force, judges the deviation of each frame rolling force setup value and actual roll-force;

S5, judge that each frame deviate is whether all in θ, θ is Deviation Control parameter, get 5%-10%, if do not meet all conditions in θ of each frame rolling force deviation, do not carry out reduction ratio study, now, in described reduction ratio learning table, other data of equivalent layer are not upgraded, and flow process enters end step;

S6, if each frame deviate, in θ, is learnt the final reduction ratio of this setting, is updated to reduction ratio study form equivalent layer by learning value other, and through learning process, this layer of other roll-force can progressively reach the requirement of target roll-force ratio;

S7, sharing of load is set and is finished.

2. fine-rolling strip steel sharing of load establishing method as claimed in claim 1, is characterized in that:

Described step S1 comprises the following steps:

S11, gets table, gets reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table, and both are added, and obtain the reduction ratio allocation table scale value of each frame

S12, carries out relativization calculating, obtains the allocation result of depressing of each frame:

Must equal the overall reduction of finish rolling according to the drafts sum of the each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{xrk}=\frac{-\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1+\sqrt{{(\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1)}^2+4\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2\&CenterDot;1\mathrm{nx}}}{2\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2}$ Formula 1

Wherein, h _nfor finish to gauge thickness, h ₀for workpiece thickness;

n is finishing pass number, and i is shelf number;

kb=0.906501，ka=0.959597。

Reduction ratio after the reduction ratio allocation table scale value relativization of each frame is the reduction ratio of depressing distribution

${\mathrm{\&epsiv;}}_{\mathrm{red}}^i=\mathrm{xrk}\&CenterDot;{\mathrm{\&epsiv;}}_0^i$ Formula 2

According to the reduction ratio of depressing distribution obtain above n-1 the frame exit thickness of depressing distribution separately

$\begin{array}{cc}{h}_{\mathrm{red}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i)& i=\mathrm{1,2},...,n-1.\end{array}$ Formula 3

3. fine-rolling strip steel sharing of load establishing method as claimed in claim 2, is characterized in that:

The reduction ratio allocation table scale value of each frame in described step S11 the process of obtaining is:

According to the large class of the current steel grade with steel, level of thickness and width class information, do not get the reduction ratio of the each frame of finish rolling from depressing distribution coefficient table equivalent layer wherein i is shelf number;

According to current tapping mark, level of thickness and width class information with steel, do not get the reduction ratio learning value of the each frame of finish rolling from reduction ratio learning table equivalent layer wherein i is shelf number, and described reduction ratio learning table is the tables of data being obtained by program automatic learning, and in table, data are constantly updated by study, and initial value gets 0;

The reduction ratio allocation table scale value of described each frame for:

4. fine-rolling strip steel sharing of load establishing method as claimed in claim 1, is characterized in that:

Described step S2 comprises the following steps:

S21, according to the large class of the current steel grade with steel and level of thickness information, does not get the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i;

S22, operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i:

Collect the operation modifying factor ζ of each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1ifor:

α _1i=α _0i(1+ ζ _i) formula 4

Wherein, ζ _ispan is-20%～20%;

Be normalized the roll-force distribution coefficient α after normalization _ifor:

S23, roll-force and the derivative of roll-force to absolute draft amount of calculating each frame under current thickness distributes:

Calculate roll-force

If: i is shelf number, and j is iterations, be the roll-force of the j time iteration of i frame, the exit thickness of the j time iteration of i frame, the exit thickness of the j time iteration of i-1 frame;

The front i frame of iterative computation exit thickness initial value for the first time the exit thickness of depressing distribution obtaining for above-mentioned formula 3

In the situation that other technological parameter is fixing, roll-force is the function of inlet thickness and exit thickness, exists: $p_i^j=p(h_{i-1}^j,h_i^j);$

Calculate the derivative of roll-force to absolute draft amount

Wherein, for the roll-force after the positive disturbance of absolute draft amount; for the roll-force after the negative disturbance of absolute draft amount, it is the absolute draft amount of the j time iteration of i frame;

There is following functional relation: $p_{i1}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\&Delta;h}}_{i1}^j),p_{i2}^j=p(h_{i-1}^j,h_{i-1}^j-{\mathrm{\&Delta;h}}_{i2}^j)$

Wherein, for the absolute draft amount after positive disturbance, for the absolute draft amount after negative disturbance,

${\mathrm{\&Delta;h}}_{i1}^j=(1+\frac{\mathrm{dt}}{2})\&CenterDot;{\mathrm{\&Delta;h}}_i^j,{\mathrm{\&Delta;h}}_{i2}^j=(1-\frac{\mathrm{dt}}{2}){\&CenterDot;\mathrm{\&Delta;h}}_i^j$

Wherein dt is disturbance, gets 1%;

S24, the result of calculation to absolute draft amount derivative according to the roll-force of above-mentioned each frame and roll-force, calculate the correction value of each frame absolute draft amount:

Rolling distributes iterative computation formula:

$\mathrm{\&delta;}\left({\mathrm{\&Delta;h}}_i^j\right)=({\mathrm{\&Sigma;p}}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{{\&PartialD;p}_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}-\mathrm{\&Sigma;}((\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{{\&PartialD;p}_i^j}{\&PartialD;\left({\mathrm{\&Delta;h}}_i^j\right)})\&CenterDot;\frac{{\mathrm{\&Delta;h}}_i^j}{\mathrm{\&Sigma;\&Delta;}{h}_i^j}$ Formula 5

Wherein: it is the correction value of the absolute draft amount of the j time iteration of i frame;

S25, the correction value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame absolute draft amount, the absolute draft amount that this renewal value is the j+1 time iteration

Computing formula is:

$\mathrm{\&Delta;}{h}_i^{j+1}=\mathrm{\&Delta;}{h}_i^j+{\mathrm{damp}}_j\&CenterDot;\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)$ Formula 6

Wherein damp _jfor damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy gets 1.0, β and gets 0.6;

S26, the renewal value of each frame absolute draft amount after obtaining, calculate the renewal value of each frame exit thickness, the exit thickness that this renewal value is the j+1 time iteration

and workpiece thickness h ₀for known quantity, according to only first frame all the other are all the inputs that previous frame is output as a rear frame, calculate last frame from first frame, obtain the renewal value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force use, wherein refer to the exit thickness of the j+1 time iteration of i-1 frame;

S27, judges whether roll-force ratio meets the condition of convergence:

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\&Sigma;}{p}_i^j}-\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}|\&le;\frac{\mathrm{\&tau;}}{{\mathrm{\&Sigma;\&alpha;}}_i}$ Formula 7

Wherein, τ is positive number, gets 0.01;

When formula (7) is set up, or iterations j while exceeding set point number finishing iteration calculate, otherwise continue to carry out described S23 step, Simultaneous Iteration number of times j cumulative 1;

S28, if roll-force ratio meets the condition of convergence in above-mentioned steps S27, judges whether the roll-force of certain frame exceedes the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

As the roll-force p of i frame _iexceed the maximum rolling force that equipment allows time, enter S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

time, carry out ${\mathrm{\&alpha;}}_i^{\&prime;}={\mathrm{\&alpha;}}_i\&CenterDot;\frac{1}{(p_i/p_i^{\max }-1)\&CenterDot;0.8+1}$ Formula 8

Wherein, α _ifor the roll-force distribution coefficient α after described normalization _i, α _i' be the roll-force distribution coefficient transfiniting after adjusting;

α _i' get back to step S23 after obtaining, with after adjusting roll-force distribution coefficient α _i' re-starting absolute draft amount iterative computation, Simultaneous Iteration number of times j assignment 0, gets the absolute draft amount of the last iteration of step S27 and exit thickness as the initial value of absolute draft amount and exit thickness;

If there is no roll-force transfinites, and directly carries out next step;

S29, absolute draft amount iterative computation and the inspection of transfiniting finish, and obtain roll-force allocation result, get the exit thickness of the last iteration of step S27 the exit thickness distributing as each frame roll-force obtain the reduction ratio that each frame roll-force is distributed simultaneously for: in formula be the exit thickness that i frame roll-force is distributed, it is the exit thickness that i-1 frame roll-force is distributed.

5. fine-rolling strip steel sharing of load establishing method as claimed in claim 1, is characterized in that:

Described step 3 comprises the following steps:

S31, carries out amplitude limiting processing to roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{xf}}^i=\left\{\begin{array}{cc}{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i& {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})& {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.$ Formula 9

Wherein for reduction ratio after amplitude limiting processing, η is amplitude limit control parameter, gets 5%～10%;

S32, to reduction ratio after amplitude limiting processing carry out relativization calculating:

If reduction ratio is modified in above-mentioned steps S31, calculating finish to gauge thickness h ' _nand between target finish to gauge thickness h n, can there is certain deviation, also will carry out taking turns relativization and calculate:

Calculating finish to gauge thickness h ' _n:

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{1n\frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after finish to gauge relativization be called final distribution reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get as the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously

$\begin{array}{cc}{h}_{\mathrm{xdh}}^i:h_{\mathrm{xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i)& i=\mathrm{1,2},...,n-1,\end{array}$

According to final distribution reduction ratio with final outlet thickness carry out the rolling procedure of fine-rolling strip steel and set calculating.

6. fine-rolling strip steel sharing of load establishing method as claimed in claim 1, is characterized in that:

In described step S6, the process that final reduction ratio is learnt is:

Ask that each frame is final distributes and reduction ratio poor of depressing distribution

${\mathrm{\&epsiv;}}_{\mathrm{dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i}{\mathrm{xrk}}$ Formula 10

Wherein, for the final reduction ratio distributing, for depressing the reduction ratio of distribution, xrk is the scale factor of the reduction ratio in formula (1);

Obtain rolling procedure and set the learning value that rear reduction ratio learning table equivalent layer does not upgrade

${\mathrm{\&epsiv;}}_{\mathrm{lrn}}^{i,\mathrm{new}}={\mathrm{\&epsiv;}}_{\mathrm{lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{dif}}^i\&CenterDot;(1-\mathrm{\&mu;})$ Formula 11

Wherein, μ is smoothing factor, gets 0.7; for rolling procedure is set the front not original learning value of reduction ratio learning table equivalent layer.