CN1850374A - Method for rolling step-thickness steel plate - Google Patents

Abstract

The present invention relates to a rolling method of step thickness steel plate. Said rolling method includes the following steps: firstly, utilizing conventional rolling process to roll the slab to larger thickness H1 of stepped plate; in the allowable range of equipment power regulating roll gap to make it meet that the outlet thickness of workpiece is smaller thickness H2 of stepped plate; after the gap regulation is completed, making low-speed rolling, rotating speed of rolls is less than 30r/min, making the rolled length of portion whose thickness is H2 be equal to L2; when the rolled length of the portion whose steel plate thickness is H2 is equal to H2, quickly lifting roll, if the step thickness difference value is large too, adopting two passes or several passes to make rolling until the required step thickness is obtained.

Description

The milling method of step-thickness steel plate

Technical field

The present invention relates to technical field of steel rolling, be specifically related to a kind of milling method of step-thickness steel plate.

Background technology

Conventional sheet material thickness in the longitudinal direction is constant substantially, but for structural member, its stressed along its length difference under many circumstances is if can produce at the different stepped plate of length direction thickness, then can effectively alleviate the structural member total amount, or reduce weld seam quantity.

Summary of the invention

At the weak point of existing step-thickness steel plate processing method, the invention provides a kind of milling method of step-thickness steel plate.

With single-order ladder thick plates is example, and technical scheme of the present invention is described:

(1) at first blank is rolled down to the big thickness H of stepped plate according to the pair rolling method ₁, its length is $L_1+L_3=L_1+\frac{H_2}{H_1}{\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="A20061004646000081.png"\; state="\{\{state\}\}">L</figure-callout>}}_2,$ As Fig. 1 (b).

(2) in the capacity of equipment allowed band, (be no more than roll-force maximum constraints and roll torque maximum constraints), adjust roller gap, make it to satisfy the less thickness H that the rolled piece exit thickness is a stepped plate ₂The method of adjustment of roll gap is as follows:

According to inlet thickness H ₁With target thickness H ₂, and information such as steel billet temperature and width, calculate rolling force F, the roll-force computing formula has a lot, and cut deal is used the Sims formula always

$F=1.15\&times;W\&CenterDot;\sqrt{R^{\&prime;}\mathrm{\&Delta;h}}\&CenterDot;Q_p\&CenterDot;\mathrm{\&sigma;}---\left(1\right)$

In the formula: W is that rolled piece width, R ' consider that roller radius, the Δ h of elastic flattening are drafts, Q _pBe that stress state influence function, σ are the average deformation drags.

After calculating roll-force according to formula (1), judge whether roll-force and roll torque surpass to allow maximum that if surpass, then be rolled with two passages, draught per pass is $\mathrm{\&Delta;H}=\frac{1}{2}(H_1-H_2).$ If calculated value is in safe range, according to the spring equation calculating gap values between rollers of formula (2).

$S=H_2-\frac{F-F_0}{K}---\left(2\right)$

In the formula: S is a gap values between rollers, mm; F ₀Be zeroing pressure, kN; K is a LONGITUDINAL STIFFNESS OF THE ROLLING MILL STAND, kN/mm.

(3) after the roll gap adjustment is finished, carry out low speed rolling, roll rotational speed generally is no more than 30r/min, makes that thickness is H ₂Part shut out length and equal L ₂, as Fig. 2.The length that shuts out part can adopt following method to calculate: on the working roll main motor shaft encoder is installed, the reading of encoder is changed the rotational speed omega that is converted into working roll, the radius R of roll is known, so according to rolling time t and advancing slip value f, just can draw the length that shuts out part.

L＝(1+f)·2πRω·t (2)

$f={\left\{\mathrm{tg}\right[\frac{\mathrm{\&pi;}}{8}\sqrt{\frac{h}{R}}\&CenterDot;\ln (1-r)+\frac{1}{2}\mathrm{arctg}\sqrt{\frac{r}{1-r}}\left]\right\}}^2$

$r=\frac{H-h}{H}$

In the formula, r is the relative reduction ratio of the operation of rolling, and H is the rolled piece inlet thickness, and h is the rolled piece exit thickness.

(4) be H at steel plate thickness ₂Part shut out length and equal L ₂The time, need lift roll fast, the mode of lifting has following two kinds:

A, hydraulic cylinder retract to the shortest position fast, depress leading screw then and also begin retraction, make roller gap greater than maximum step-thickness H ₁, last roll and the revolution of front and back roller-way are withdrawed from roll after making rolled piece, as Fig. 3;

B, hydraulic cylinder bounce back fast, depress leading screw then and also begin retraction, make roller gap greater than maximum step-thickness H ₁, roll and front and back roller-way stop operating, and make rolled piece be parked in original position, treat the roller gap stable reading after, restart roll and front and back roller-way, gradually rolled piece is transported out milling train;

(5) if in the step (2), step-thickness difference DELTA H=H ₁-H ₂Too big, cause roll-force or roll torque too big easily with a passage, need to adopt two passages or more multi-pass be rolled, then carry out the operation of step (2) later on again to (4) in a rolling passage, can obtain the step-thickness that needs.

Remarkable result of the present invention is: can be with technology conveniently, and continuous rolling is produced the step-thickness plate.

Description of drawings

Fig. 1 is a step-thickness board size schematic diagram, (a) size schematic diagram after rolling wherein, the size schematic diagram before (b) rolling;

Fig. 2 is an operation of rolling schematic diagram;

Fig. 3 is a milling train pressing member schematic diagram, and 1 depresses leading screw among the figure, 2 hydraulic cylinders, 3 pressure sensors, 4 backing rolls, 5 working rolls, 6 rolled pieces.

The specific embodiment

Embodiment 1

The blank of dimensional thickness 220mm * wide 1600mm * length 2600, roll into the step-thickness plate of 30mm * 2600mm * 6000mm-28mm * 2600mm * 6100mm by requirement shown in Figure 1, LONGITUDINAL STIFFNESS OF THE ROLLING MILL STAND is 10000kN/mm, zeroing pressure 30000kN, roller radius 500mm, maximum permission roll-force 55000kN, maximum permission roll torque 5000kNm.

At first with 12 passages rolled piece is rolled into the steel plate of 30mm * 2600mm * 11733mm according to the pair rolling method, calculates rolling force F then, inlet thickness is 30mm, the target exit thickness is 28mm, corresponding roll-force is 12500kN, and roll torque 395kNm is in safe range.

Calculate fixed value of roller slit S

$S=H_2-\frac{F-F_0}{K}=28-\frac{12500-30000}{10000}=28+1.75=29.75$

Additional then passage is adjusted the roller gap setting value, makes it to equal 29.75mm, makes the speed at the uniform velocity steel rolling of roll with 20r/m then, because the length of the thick correspondence of steel plate 28mm is 6100mm, can calculate its corresponding rolling time according to formula (3)

$t=\frac{L}{(1+f)\&CenterDot;2\mathrm{\&pi;R\&omega;}}=\frac{6100}{(1+0.016)\&times;2\&times;3.1416\&times;500}\&times;3=5.733s$

When rolling time equals 5.733s, make the hydraulic cylinder retraction fast, lift roll then, and make roll and roller-way counter-rotating, make rolled piece withdraw from milling train, then rolled piece is just rolled into the step-thickness plate.

Embodiment 2

The blank of dimensional thickness 220mm * wide 1600mm * length 2600, roll into the step-thickness plate of 30mm * 2600mm * 6000mm-20mm * 2600mm * 8600mm by requirement shown in Figure 1, LONGITUDINAL STIFFNESS OF THE ROLLING MILL STAND is 10000kN/mm, zeroing pressure 30000kN, roller radius 500mm, maximum permission roll-force 55000kN, maximum permission roll torque 5000kNm.。

At first with 12 passages rolled piece is rolled into the steel plate of 30mm * 2600mm * 11733mm according to the pair rolling method, calculates rolling force F then, inlet thickness is 30mm, the target exit thickness is 20mm, corresponding roll-force is 62500kN, roll torque 4419kNm, and roll-force surpasses safe range.So need to adopt a plurality of passages to be rolled, suppose first passage with steel plate rolling 25mm, corresponding roll-force 25000kN, roll torque 1250kNm, roll-force and roll torque are all in safe range.

Calculate fixed value of roller slit S

$S=H_2-\frac{F-F_0}{K}=25-\frac{25000-30000}{10000}=25+0.5=25.5$

Additional then passage is adjusted the roller gap setting value, makes it to equal 25.5mm, makes the speed at the uniform velocity steel rolling of roll with 20r/m then, because the length of the thick correspondence of steel plate 25mm is 6880mm, can calculate its corresponding rolling time according to formula (3)

$t=\frac{L}{(1+f)\&CenterDot;2\mathrm{\&pi;R\&omega;}}=\frac{6880}{(1+0.039)\&times;2\&times;3.1416\&times;500}\&times;3=6.323s$

When rolling time equals 6.323s, make the hydraulic cylinder retraction fast, lift roll then, and make roll and roller-way counter-rotating, make rolled piece withdraw from milling train.

Calculate the roll-force and the roll torque of second passage then, inlet thickness is 25mm, and the target exit thickness is 20mm, and roll-force is 37500kN, and roll torque is 1875kNm.Roll-force and roll torque are all in safe range.

Calculate fixed value of roller slit S

$S={\mathrm{<figure-callout\; id="2"\; label="H"\; filenames="A20061004646000081.png"\; state="\{\{state\}\}">H</figure-callout>}}_2^{\&prime;}-\frac{F^{\&prime;}-F_0}{K}=20-\frac{37500-30000}{10000}=20-0.75=19.25$

Additional then passage is adjusted the roller gap setting value, makes it to equal 19.25mm, makes the speed at the uniform velocity steel rolling of roll with 20r/m then, because the length of the thick correspondence of steel plate 20mm is 8600mm, can calculate its corresponding rolling time according to formula (3)

$t=\frac{L}{(1+f)\&CenterDot;2\mathrm{\&pi;R\&omega;}}=\frac{8600}{(1+0.047)\&times;2\&times;3.1416\&times;500}\&times;3=7.844s$

When rolling time equals 7.844s, make the hydraulic cylinder retraction fast, lift roll then, and make roll and roller-way counter-rotating, make rolled piece withdraw from milling train.

Embodiment 3

The blank of dimensional thickness 220mm * wide 1400mm * length 2200, roll into the step-thickness plate of 18mm * 2500mm * 6000mm-15mm * 2500mm * 10190mm by requirement shown in Figure 1, LONGITUDINAL STIFFNESS OF THE ROLLING MILL STAND is 6000kN/mm, zeroing pressure 15000kN, roller radius 430mm, maximum permission roll-force 35000kN, maximum permission roll torque 2500kNm.

At first with 12 passages rolled piece is rolled into the steel plate of 18mm * 2500mm * 15058mm according to the pair rolling method, calculates rolling force F then, inlet thickness is 18mm, the target exit thickness is 15mm, corresponding roll-force is 21500kN, and roll torque 772kNm is in safe range.

Calculate fixed value of roller slit S

$S=H_2-\frac{F-F_0}{K}=18-\frac{21500-15000}{6000}=18-1.083=16.917$

Additional then passage is adjusted the roller gap setting value, makes it to equal 16.917mm, makes the speed at the uniform velocity steel rolling of roll with 20r/m then, because the length of the thick correspondence of steel plate 15mm is 10190mm, can calculate its corresponding rolling time according to formula (3)

$t=\frac{L}{(1+f)\&CenterDot;2\mathrm{\&pi;R\&omega;}}=\frac{10190}{(1+0.040)\&times;2\&times;3.1416\&times;430}\&times;3=10.880s$

When rolling time equals 10.880s, make the hydraulic cylinder retraction fast, lift roll then, and make roll and roller-way counter-rotating, make rolled piece withdraw from milling train, then rolled piece is just rolled into the step-thickness plate.

Claims (4)

1, a kind of milling method of step-thickness steel plate is characterized in that carrying out according to the following steps:

(1) at first blank is rolled down to the big thickness H of stepped plate according to the pair rolling method ₁

(2) in the capacity of equipment allowed band, promptly be no more than roll-force maximum constraints and roll torque maximum constraints, adjust roller gap, make it to satisfy the less thickness H that the rolled piece exit thickness is a stepped plate ₂

(3) after the roll gap adjustment is finished, carry out low speed rolling, roll rotational speed is no more than 30r/min, makes that thickness is H ₂Part shut out length and equal L ₂

(4) be H at steel plate thickness ₂Part shut out length and equal L ₂The time, lift roll fast;

(5) if in the step (2), step-thickness difference DELTA H=H ₁-H ₂Surpass roll-force maximum constraints and roll torque maximum constraints, then adopt two passages or more multi-pass be rolled, carry out the operation of step (2) later on again in a rolling passage, until the step-thickness that needing to obtain to (4).

2,, it is characterized in that the method for adjustment of roll gap in the step (2) is as follows according to the milling method of the described step-thickness steel plate of claim 1:

According to inlet thickness H ₁With target thickness H ₂, and steel billet temperature and width information, calculate rolling force F, judge whether roll-force and roll torque surpass the permission maximum, if surpass, then are rolled with two or more passages, and draught per pass is $\mathrm{\&Delta;H}=\frac{1}{n}(H_1-H_2),$ If calculated value in safe range, calculates gap values between rollers according to the spring equation of following formula,

$S=H_2-\frac{F-F_0}{K}$

In the formula: S is a gap values between rollers, mm; F ₀Be zeroing pressure, kN; K is a LONGITUDINAL STIFFNESS OF THE ROLLING MILL STAND, and kN/mm, n are number of times.

3,, it is characterized in that the length that shuts out part in the step (3) adopts following method to calculate according to the milling method of the described step-thickness steel plate of claim 1:

On the working roll main motor shaft encoder is installed, the reading of encoder is changed the rotational speed omega that is converted into working roll, the radius R of roll is known, according to rolling time t and advancing slip value f, calculates the length that shuts out part,

L＝(1+f)·2πRω·t

$f={\left\{\mathrm{tg}\right[\frac{\mathrm{\&pi;}}{8}\sqrt{\frac{h}{R}}\&CenterDot;\ln (1-r)+\frac{1}{2}\mathrm{arctg}\sqrt{\frac{r}{1-r}}\left]\right\}}^2$

$r=\frac{H-h}{H}$

In the formula, r is the relative reduction ratio of the operation of rolling, and H is the rolled piece inlet thickness, and h is the rolled piece exit thickness.

4,, it is characterized in that lifting in the step (4) the roll mode and adopt following a kind of according to the milling method of the described step-thickness steel plate of claim 1:

A, hydraulic cylinder retract to the shortest position fast, depress leading screw then and also begin retraction, make roller gap greater than maximum step-thickness H ₁, last roll and the revolution of front and back roller-way are withdrawed from roll after making rolled piece;

B, hydraulic cylinder bounce back fast, depress leading screw then and also begin retraction, make roller gap greater than maximum step-thickness H ₁, roll and front and back roller-way stop operating, and make rolled piece be parked in original position, treat the roller gap stable reading after, restart roll and front and back roller-way, gradually rolled piece is transported out milling train.

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