CN101566538B - On-line acquisition method for plasticity coefficient of rolled piece during rolling of medium plate - Google Patents

Abstract

The invention relates to an on-line acquisition method for plasticity coefficient of a rolled piece during rolling of a medium plate, and belongs to the technical field of rolling. The method comprises the following steps: (1) determining the inlet thickness of the rolled piece; (2) acquiring the actual outlet thickness of the rolled piece; (3) solving key points on a plastic curve; (4) fitting the plastic curve; (5) calculating the tangent slope at the actual pressing measurement point to obtain the plasticity coefficient; (6) applying the solved plasticity coefficient to an AGC control modelon line; and (7) triggering next period, turning to step (1) and reacquiring the plasticity coefficient according to the acquired data. The method has the advantages that the method does not depend o n the investment of process computers and is not influenced by complex factors in a production field; the result has no jump phenomenon; and the acquisition process is stable, can be directly embeddedinto basic automation to be applied, and continuously correct the plasticity coefficient along with the rolling process according to the change of a roll force and the change of a roll gap so as to i mprove the thickness compensation precision of an AGC system and apply to high-precision AGC control.

Description

On-line acquisition method for plasticity coefficient of rolled piece in a kind of Medium and Heavy Plate Rolling process

Technical field:

The invention belongs to rolling technical field, on-line acquisition method for plasticity coefficient of rolled piece in particularly a kind of Medium and Heavy Plate Rolling process.

Background technology:

At present, heavy and medium plate mill generally adopts automatic gauge control system in the operation of rolling, is called for short AGC, and AGC controlling models commonly used is thickness meter formula AGC controlling models and dynamically sets the AGC model that its model form is shown in (1) and (2).

Thickness meter AGC model: $\mathrm{\Δ}{S}_k=-\frac{M+Q}{M}\·\mathrm{\Δ}{h}_k---\left(1\right)$

Dynamically set the AGC model: $\mathrm{\Δ}{S}_k=-(\frac{Q}{M}\·\mathrm{\Δ}{S}_{k-1}+\frac{M+Q}{M^2}\·\mathrm{\Δ}{P}_k)---\left(2\right)$

In the formula: M is the mill stiffness curve; Q is a plasticity coefficient of rolled piece; Δ S _K-1Be a last moment roll gap regulated quantity.

In the AGC controlling models, mill stiffness M and plasticity coefficient of rolled piece Q are major parameter, directly have influence on thickness, the compensation precision of rolled piece, mill stiffness can press the spring curve by the full body of roll that obtains when the rigidity test, and the rolled piece width compensated and precisely controlled, required roll-force when plasticity coefficient of rolled piece Q is defined as rolled piece and produces unit deformation, that is:

$Q=-\frac{\∂P}{\∂h}---\left(3\right)$

Q is a rolled piece mechanical property parameters in the reaction operation of rolling, at present in the AGC system, obtaining mainly of plasticity coefficient of rolled piece finished by process computer, according to rolling schedule by formula P/ (H-h) for every time draws an approximate coefficient of plasticity, before stinging steel, send to basic automatization as rolled piece AGC controlled variable.Adopt this method that following shortcoming is arranged:

1. it is big to obtain error, and the tangent slope and the average gradient difference at promptly actual drafts point place are big;

2. the operation of basic automatization AGC function relies on process computer, and when the process computing machine was shut down, basic automatization can only adopt default value;

3. in the operation of rolling, rolled piece is influenced by the head and the tail temperature difference, watermark extraneous factor, and the steel plate thickness fluctuation directly influences the actual value of plasticity coefficient of rolled piece, uses a fixing average coefficient of plasticity to carry out the AGC model and tries to achieve, and can cause system's THICKNESS CONTROL inaccurate.

Summary of the invention:

The deficiency of obtaining and use existence at existing milling train plasticity coefficient of rolled piece in the operation of rolling, the invention provides on-line acquisition method for plasticity coefficient of rolled piece in a kind of Medium and Heavy Plate Rolling process, based on rolling mechanism model, utilize secondary or curve fitting plasticity curve repeatedly, can draw the tangent slope at drafts point place easily, be coefficient of plasticity, process computer is not adopted in obtaining of coefficient of plasticity, directly embed to move in the basic automatization and obtain, before obtaining the AGC regulated quantity, obtains one time coefficient of plasticity in each cycle, like this along with the carrying out of the operation of rolling, according to the variation of roll-force and the variation of roll gap, constantly coefficient of plasticity is revised, thus the thickness compensation precision of raising AGC system;

Effective for guaranteeing AGC employed coefficient of plasticity of each cycle of control system, should before implementing, AGC carry out trying to achieve of coefficient of plasticity.

On-line acquisition method for plasticity coefficient of rolled piece in a kind of Medium and Heavy Plate Rolling process of the present invention, step is as follows:

1. determine the rolled piece inlet thickness

For first passage, inlet thickness H is the thickness of supplied materials, and for other passage, inlet thickness adopts formula (1) to calculate:

H＝gap _avg+(P _avg-P _zero)/M+W (1)

In the formula, gap _AvgBe to go up the average roll gap of passage, p _AvgBe to go up the average roll-force of passage; M is a mill stiffness of considering dimension of roller and rolled piece width compensation, and W is a roll wear, uses formula (2) to calculate:

$W=\underset{n}{\mathrm{\Σ}}a{A}^{\mathrm{\α}}{B}^{\mathrm{\β}}C---\left(2\right)$

Wherein: A is that load value influences item, and B is a contact arc length influence item, and C is a mill length influence item, and a, α, β are regression coefficients, and n is a rolling pass;

2. obtain actual rolled piece exit thickness according to production scene image data and wear model

The actual exit thickness of rolled piece is calculated by formula (3):

h _act＝gap _act+(P _act-P _zero)/M+W (3)

In the formula, h _ActIt is the actual exit thickness of rolled piece in the operation of rolling; Gap _ActIt is actual roll gap; P _ActIt is actual roll-force; P _ZeroIt is the zero clearing roll-force;

3. try to achieve key point on the plasticity curve by rolling mechanism model

The roll-force computing formula adopts the SIMS formula in the course of hot rolling:

P＝1.15·σ·Q _P·l _C·B (4)

In the formula: σ is a resistance of deformation; Q _PBe distorted area shape influence function; l _CBe the arc of contact length of considering elastic flattening; B is the rolled piece width;

Deformation resistance model adopts the good type structure that helps of U.S. slope, is defined as following form:

$\mathrm{\σ}=a_0\·\exp (a_1+a_{2}T)\·{\mathrm{\ϵ}}^{a_3}\·{\stackrel{\·}{\mathrm{\ϵ}}}^{a_4}---\left(5\right)$

In the formula: T=(t °+273)/1000, ε is strain,

Be strain rate, a ₀～a ₄It is the regression coefficient of corresponding different steel grades;

The plasticity curve has reflected the relation between rolled piece thickness and the roll-force, for using secondary or curve fitting plasticity curve repeatedly, need obtain key point on the plasticity curve, and wherein key point is the required data point of match quafric curve, and known key point has two:

(H, 0) and (h _Act, P _Act)

Wherein H is an inlet thickness; h _ActBe the actual exit thickness of rolled piece in the operation of rolling, calculate by formula (3);

Suppose that strain is ε in the operation of rolling, according to formula (4), can obtain the roll-force P (α ε) when strain is α ε (0＜α＜1), P (ε) is shown below with P (α ε) ratio:

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)Q_P\left(\mathrm{\ϵ}\right)l_c\left(\mathrm{\ϵ}\right)B\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)Q_P\left(\mathrm{\α\ϵ}\right)l_c\left(\mathrm{\α\ϵ}\right)B\left(\mathrm{\α\ϵ}\right)}---\left(6\right)$

Distorted area shape influence function Q in the operation of rolling _pVery little with the variation of rolled piece width B, can ignore, formula (6) is reduced to following form:

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)l_c\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)l_c\left(\mathrm{\α\ϵ}\right)}---\left(7\right)$

Strain stress is represented with inlet, exit thickness:

$\mathrm{\ϵ}=\ln \left(\frac{H}{h}\right)---\left(8\right)$

In strain is that drafts ratio is under ε and the α ε:

$\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}=e^{\mathrm{\α\ϵ}-\mathrm{\ϵ}}\frac{{\mathrm{He}}^{\mathrm{\ϵ}}-H}{{\mathrm{He}}^{\mathrm{\α\ϵ}}-H}=e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}---\left(9\right)$

In strain is that the ratio of resistance of deformation is under ε and the α ε:

$\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)}=\frac{{\mathrm{\ϵ}}^{a_3+a_4}\·{\left(\frac{\sqrt{\mathrm{R\Δ}{h}_1}}{2\mathrm{\πRn}/60}\right)}^{-a_4}}{{\left(\mathrm{\α\ϵ}\right)}^{a_3+a_4}\·{\left(\frac{\sqrt{\mathrm{R\Δ}{h}_2}}{2\mathrm{\πRn}/60}\right)}^{-a_4}}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}\right)}^{-\frac{a_4}{2}}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{-\frac{a_4}{2}}---\left(10\right)$

In strain is that the ratio of arc of contact length is under ε and the α ε:

$\frac{L_C\left(\mathrm{\ϵ}\right)}{L_C\left(\mathrm{\α\ϵ}\right)}=\sqrt{\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}}=\sqrt{\frac{H-\frac{H}{e^{\mathrm{\ϵ}}}}{H-\frac{H}{e^{\mathrm{\α\ϵ}}}}}={\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{n\ϵ}}-1}\right)}^{1/2}---\left(11\right)$

To obtain in formula (10), (11) substitution formula (7):

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}}---\left(12\right)$

In strain be for α ε place draught pressure:

$P\left(\mathrm{\α\ϵ}\right)=P\left(\mathrm{\ϵ}\right){\mathrm{\α}}^{a_3+a_4}/{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}}---\left(13\right)$

By formula (13) as can be known, when α gets different value, can try to achieve P (α ε), promptly obtain required other key point (h (α ε), P (α ε)) of matched curve, that is: by P (ε)

$(H/e^{\mathrm{\α\ϵ}},P\left(\mathrm{\ϵ}\right){\mathrm{\α}}^{a_3+a_4}/{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}});$

4. according to obtaining key point, utilize secondary or curve fitting plasticity curve repeatedly

If use conic fitting plasticity curve, get α=0.5, calculate the 3rd required key point of match quafric curve (h (0.5 ε), P (0.5 ε)) by following formula, suppose that the form of quafric curve is as follows:

y＝b ₀+b ₁x+b ₂x ² (14)

With (H, 0), (h _Act, P _Act) and (h (0.5 ε), P (0.5 ε)) substitution can obtain b ₀, b ₁And b ₂

5. on the curve of match, calculate the tangent slope at actual drafts point place, obtain coefficient of plasticity in quafric curve (14), the parameter b of trying to achieve ₀, b ₁And b ₂As follows:

$b_2=\frac{P_{\mathrm{act}}(h\left(0.5\mathrm{\ϵ}\right)-H)-P\left(0.5\mathrm{\ϵ}\right)(h_{\mathrm{act}}-H)}{(h\left(0.5\mathrm{\ϵ}\right)-H)(h_{\mathrm{act}}-H)(h_{\mathrm{act}}-h\left(0.5\mathrm{\ϵ}\right))}$

$b_1=\frac{P_{\mathrm{act}}}{h_{\mathrm{act}}-H}-b_2(h_{\mathrm{act}}+H)---\left(15\right)$

b ₀＝-b ₁H-b ₂H ²

Coefficient of plasticity is quafric curve at actual drafts point (h _Act, P _Act) tangent slope located, that is:

Q＝y′(h _act)＝b ₁+2b ₂h _act (16)

6. with in the online AGC of the being applied to controlling models of the coefficient of plasticity of trying to achieve;

7. the next cycle triggers, and changes step over to and 1. obtains coefficient of plasticity again according to image data;

Advantage of the present invention:

The on-line acquisition method for plasticity coefficient of deriving based on rolling mechanism model, the process formula that obtains coefficient of plasticity is simple, do not rely on the input of process computer, be not subjected to the influence of production scene complicated factor, the result does not have the saltus step phenomenon, acquisition process is stable, can directly embed in the basic automatization and use, and carrying out along with the operation of rolling, according to the variation of roll-force and the variation of roll gap, constantly coefficient of plasticity is revised, thereby the thickness compensation precision of raising AGC system is applicable to high-precision A GC control; The coefficient of plasticity that the present invention obtains is the tangent slope at actual drafts point place on the plasticity curve, fits like a glove with the definition of coefficient of plasticity, and the precision height can improve solving precision more than 6% than the classic method of finding the solution coefficient of plasticity mean value.

Description of drawings:

On-line acquisition method for plasticity coefficient of rolled piece process flow diagram in a kind of Medium and Heavy Plate Rolling process of Fig. 1 the present invention;

Plasticity coefficient of rolled piece is used AGC control block diagram in a kind of Medium and Heavy Plate Rolling process of Fig. 2 the present invention.

Embodiment:

The on-line acquisition method for plasticity coefficient of rolled piece detailed process is illustrated in conjunction with the embodiments in a kind of Medium and Heavy Plate Rolling process of the present invention.Plasticity coefficient of rolled piece is used AGC control block diagram as shown in Figure 2 in a kind of Medium and Heavy Plate Rolling process.

Present embodiment is selected the Q235 steel grade, and parameter is as follows:

Steel grade: Q235

Blank specification: 220mm * 1600mm * 2810mm

Finished size: 14mm * 2120mm

Tapping temperature: 1100 ℃

Rolling pass: 15 passages

Mill stiffness M:8000KN/mm

Zero clearing roll-force: 18560KN/mm

Resistance of deformation parameter: a ₃=0.000027, a ₄=0.25381

Rolling procedure is set as shown in table 1,

Table 1 rolling procedure table

The present invention illustrates on-line acquisition method for plasticity coefficient of rolled piece in the Medium and Heavy Plate Rolling process in conjunction with the embodiments, and step is as follows: as shown in Figure 1,

1. determine the rolled piece inlet thickness

The inlet thickness of the 1st passage is supplied materials thickness H=220mm;

2. obtain actual rolled piece exit thickness according to production scene image data and wear model

After stinging steel, PLC arrives actual roll gap gap in certain cycle detection ₁Be 204.4mm, actual roll-force P ₁=26784KN, be 0.0124mm by the roll wear W that formula (2) calculates this moment, can calculate rolled piece exit thickness h by formula (3) ₁:

h ₁＝204.4+(26784-18560)/8000+0.0124＝205.4405mm

3. try to achieve key point on the plasticity curve by rolling mechanism model

For steel grade Q235, the resistance of deformation coefficient is as follows:

a ₃＝2.7×10 ^-5，a ₄＝0.2538

Utilize formula (13) to ask the 3rd some h (0.5 ε) and P (0.5 ε) on the plasticity curve:

h ₂＝h(0.5ε)＝H/e ^0.5ε＝212.59565mm

$P_2=P\left(0.5\mathrm{\ϵ}\right)=P\left(\mathrm{\ϵ}\right){0.5}^{a3+a4}/{\left(e^{(0.5-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{0.5\mathrm{\ϵ}}-1}\right)}^{\frac{1-a_4}{2}}=17454.2024\mathrm{KN};$

Three key points that obtain on the plasticity curve are as follows:

(220，0)，(205.4405，26784)，(212.59565，17454.2024)；

4. according to obtaining key point, utilize conic fitting plasticity curve

Obtain the parameter of quafric curve by formula (15):

b ₀＝-2.865×10 ⁶，b ₁＝2.894×10 ⁴，b ₂＝-72.35；

5. on the curve of match, calculate the tangent slope at actual drafts point place, obtain coefficient of plasticity at (h ₁, P ₁) the moment, the coefficient of plasticity that is calculated by formula (16) is:

Q＝786.2614KN/mm；

6. with in the online AGC of the being applied to controlling models of the coefficient of plasticity of trying to achieve;

7. the next cycle triggers, and changes step over to and 1. obtains coefficient of plasticity again according to image data;

When roll gap that collects following one-period and roll-force data, try to achieve according to above step equally.For other passage beyond first passage, the average roll gap of passage and average roll-force adopt formula (1) to try to achieve inlet thickness in the utilization, and other step of obtaining coefficient of plasticity is identical.

The present invention is applicable to heavy and medium plate mill, equally also the roughing mill of applied heat continuous rolling process and finishing mill.

Claims (1)

1. on-line acquisition method for plasticity coefficient of rolled piece in the Medium and Heavy Plate Rolling process, it is characterized in that: this method step is as follows:

1. determine the rolled piece inlet thickness

For first passage, inlet thickness H is the thickness of supplied materials, and for other passage, inlet thickness adopts formula (1) to calculate:

H＝gap _avg+(P _avg-P _zero)/M+W (1)

In the formula, gap _AvgBe to go up the average roll gap of passage, p _AvgBe to go up the average roll-force of passage; M is a mill stiffness of considering dimension of roller and rolled piece width compensation, and W is a roll wear, P _ZeroIt is the zero clearing roll-force; Use formula (2) to calculate:

$W=\underset{n}{\mathrm{\Σ}}a{A}^{\mathrm{\α}}{B}^{\mathrm{\β}}C---\left(2\right)$

Wherein: A is that load value influences item, and B is a contact arc length influence item, and C is a mill length influence item, and a, α, β are regression coefficients, and n is a rolling pass;

2. obtain actual rolled piece exit thickness according to production scene image data and wear model

The actual exit thickness of rolled piece is calculated by formula (3):

h _act＝gap _act+(P _act-P _zero)/M+W (3)

In the formula, h _ActIt is the actual exit thickness of rolled piece in the operation of rolling; Gap _ActIt is actual roll gap; P _ActIt is actual roll-force; P _ZeroIt is the zero clearing roll-force; M is a mill stiffness of considering dimension of roller and rolled piece width compensation; W is a roll wear;

3. try to achieve key point on the plasticity curve by rolling mechanism model

Select roll-force computing formula and deformation resistance model

The roll-force computing formula of course of hot rolling adopts the SIMS formula:

P＝1.15·σ·Q _P·l _C·B (4)

In the formula: σ is a resistance of deformation; Q _PBe distorted area shape influence function; l _CBe the arc of contact length of considering elastic flattening; B is the rolled piece width;

Deformation resistance model adopts the good type structure that helps of U.S. slope, is defined as following form:

$\mathrm{\σ}=a_0\·\exp (a_1+a_{2}T)\·{\mathrm{\ϵ}}^{a_3}\·{\stackrel{\·}{\mathrm{\ϵ}}}^{a_4}---\left(5\right)$

In the formula: T=(t °+273)/1000, ε is strain,

Be strain rate, a ₀～a ₄Be the regression coefficient of corresponding different steel grades, t ° is the actual Celsius temperature of rolled piece in the operation of rolling;

The plasticity curve has reflected the relation between rolled piece thickness and the roll-force, for using secondary or curve fitting plasticity curve repeatedly, need obtain key point on the plasticity curve, and known key point has two:

(H, 0) and (h _Act, P _Act)

Wherein H is an inlet thickness; h _ActBe the actual exit thickness of rolled piece in the operation of rolling, P _ActRepresent actual roll-force; Calculate by formula (3);

Suppose that strain is ε in the operation of rolling,, can obtain the roll-force P (α ε) when strain is α ε according to formula (4), the roll-force when P (ε) is ε for strain, 0＜α＜1 wherein, P (ε) is shown below with P (α ε) ratio:

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)Q_P\left(\mathrm{\ϵ}\right)l_c\left(\mathrm{\ϵ}\right)B\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)Q_P\left(\mathrm{\α\ϵ}\right)l_c\left(\mathrm{\α\ϵ}\right)B\left(\mathrm{\α\ϵ}\right)}---\left(6\right)$

Distorted area shape influence function Q in the operation of rolling _pVery little with the variation of rolled piece width B, can ignore, formula (6) is reduced to following form:

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)l_c\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)l_c\left(\mathrm{\α\ϵ}\right)}---\left(7\right)$

Strain stress is represented with inlet, exit thickness:

$\mathrm{\ϵ}=\ln \left(\frac{H}{h}\right)---\left(8\right)$

Wherein H represents inlet thickness, and h represents exit thickness;

In strain is that drafts ratio is under ε and the α ε:

$\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}=e^{\mathrm{\α\ϵ}-\mathrm{\ϵ}}\frac{H{e}^{\mathrm{\ϵ}}-H}{{\mathrm{He}}^{\mathrm{\α\ϵ}}-H}=e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}---\left(9\right)$

In strain is that the ratio of resistance of deformation is under ε and the α ε:

$\frac{\mathrm{\σ}\left(\mathrm{\ϵ}\right)}{\mathrm{\σ}\left(\mathrm{\α\ϵ}\right)}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}\right)}^{-\frac{a_4}{2}}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{-\frac{a_4}{2}}---\left(10\right)$

In strain is that the ratio of arc of contact length is under ε and the α ε:

$\frac{L_C\left(\mathrm{\ϵ}\right)}{L_C\left(\mathrm{\α\ϵ}\right)}=\sqrt{\frac{\mathrm{\Δh}\left(\mathrm{\ϵ}\right)}{\mathrm{\Δh}\left(\mathrm{\α\ϵ}\right)}}=\sqrt{\frac{H-\frac{H}{e^{\mathrm{\ϵ}}}}{H-\frac{H}{e^{\mathrm{\α\ϵ}}}}}={\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{1/2}---\left(11\right)$

To obtain in formula (10), (11) substitution formula (7):

$\frac{P\left(\mathrm{\ϵ}\right)}{P\left(\mathrm{\α\ϵ}\right)}=\frac{1}{{\mathrm{\α}}^{a_3+a_4}}\·{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}}---\left(12\right)$

In strain be for α ε place draught pressure:

$P\left(\mathrm{\α\ϵ}\right)=P\left(\mathrm{\ϵ}\right){\mathrm{\α}}^{a_3+a_4}/{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}}---\left(13\right)$

By formula (13) as can be known, when α gets different value, can try to achieve P (α ε), promptly obtain required other key point (h (α ε), P (α ε)) of matched curve, that is: by P (ε)

$(H/e^{\mathrm{\α\ϵ}},P\left(\mathrm{\ϵ}\right){\mathrm{\α}}^{a_3+a_4}/{\left(e^{(\mathrm{\α}-1)\mathrm{\ϵ}}\frac{e^{\mathrm{\ϵ}}-1}{e^{\mathrm{\α\ϵ}}-1}\right)}^{\frac{1-a_4}{2}});$

4. according to the key point that obtains, utilize secondary or curve fitting plasticity curve repeatedly

If use conic fitting plasticity curve, get α=0.5, calculate the 3rd required key point of match quafric curve (h (0.5 ε), P (0.5 ε)) by following formula, suppose that the form of quafric curve is as follows:

y＝b ₀+b ₁x+b ₂x ² (14)

With (H, 0), (h _Act, P _Act) and (h (0.5 ε), P (0.5 ε)) substitution can obtain b ₀, b ₁And b ₂

5. on the curve of match, calculate the tangent slope at actual drafts point place, obtain coefficient of plasticity

In quafric curve (14), the parameter b of trying to achieve ₀, b ₁And b ₂As follows:

$b_2=\frac{P_{\mathrm{act}}(h\left(0.5\mathrm{\ϵ}\right)-H)-P\left(0.5\mathrm{\ϵ}\right)(h_{\mathrm{act}}-H)}{(h\left(0.5\mathrm{\ϵ}\right)-H)(h_{\mathrm{act}}-H)(h_{\mathrm{act}}-h\left(0.5\mathrm{\ϵ}\right))}$

$b_1=\frac{P_{\mathrm{act}}}{h_{\mathrm{act}}-H}-b_2(h_{\mathrm{act}}+H)---\left(15\right)$

b ₀＝-b ₁H-b ₂H ²

Coefficient of plasticity is quafric curve at actual drafts point (h _Act, P _Act) tangent slope located, that is:

Q＝y′(h _act)＝b ₁+2b ₂h _act (16)

6. with in the online AGC of the being applied to controlling models of the coefficient of plasticity of trying to achieve;

7. the next cycle triggers, and changes step over to and 1. obtains coefficient of plasticity again according to image data.

Images