ROLL PASS DESIGN

IN CONTINUOUS

BAR MILLS

Department of Metallurgical

and Materials Engineering

INDIAN INSTITUTE OF

TECHNOLOGY

KHARAGPUR

bullTwo facing grooves form a lsquoroll passrsquo or simply a

lsquopassrsquo The distance between the barrels of two rolls is

called the lsquonominal roll gaprsquo or lsquotheoretical roll gaprsquo

bullThe ultimate goal of a roll pass design is to ensure the

production of the desired shape of a product with the

appropriate internal structure defect free surface and

at lowest cost

Basic Terminology

The basic five different cross-section

shapes used in roll pass design

Passes amp Bars

bullDefinite passes ndash those

having two equal axes in

an x y plane (Squares

Rounds)

bullIntermediate passes ndash

those having one axis

larger than the other one

(Rectangles ndash box

Diamonds Ovals)

Deformation amp Sequence

bullA definite bar into one intermediate pass or an intermediate bar into one definite pass configures a lsquodeformationrsquo For example a square into an oval pass or an oval into a square pass A deformation can produce any type of bar bullA definite bar into two passes (an intermediate pass followed by a definite pass configures a lsquosequencersquo A sequence only produces a definite bar

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullTwo facing grooves form a lsquoroll passrsquo or simply a

lsquopassrsquo The distance between the barrels of two rolls is

called the lsquonominal roll gaprsquo or lsquotheoretical roll gaprsquo

bullThe ultimate goal of a roll pass design is to ensure the

production of the desired shape of a product with the

appropriate internal structure defect free surface and

at lowest cost

Basic Terminology

The basic five different cross-section

shapes used in roll pass design

Passes amp Bars

bullDefinite passes ndash those

having two equal axes in

an x y plane (Squares

Rounds)

bullIntermediate passes ndash

those having one axis

larger than the other one

(Rectangles ndash box

Diamonds Ovals)

Deformation amp Sequence

bullA definite bar into one intermediate pass or an intermediate bar into one definite pass configures a lsquodeformationrsquo For example a square into an oval pass or an oval into a square pass A deformation can produce any type of bar bullA definite bar into two passes (an intermediate pass followed by a definite pass configures a lsquosequencersquo A sequence only produces a definite bar

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

The basic five different cross-section

shapes used in roll pass design

Passes amp Bars

bullDefinite passes ndash those

having two equal axes in

an x y plane (Squares

Rounds)

bullIntermediate passes ndash

those having one axis

larger than the other one

(Rectangles ndash box

Diamonds Ovals)

Deformation amp Sequence

bullA definite bar into one intermediate pass or an intermediate bar into one definite pass configures a lsquodeformationrsquo For example a square into an oval pass or an oval into a square pass A deformation can produce any type of bar bullA definite bar into two passes (an intermediate pass followed by a definite pass configures a lsquosequencersquo A sequence only produces a definite bar

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Passes amp Bars

bullDefinite passes ndash those

having two equal axes in

an x y plane (Squares

Rounds)

bullIntermediate passes ndash

those having one axis

larger than the other one

(Rectangles ndash box

Diamonds Ovals)

Deformation amp Sequence

bullA definite bar into one intermediate pass or an intermediate bar into one definite pass configures a lsquodeformationrsquo For example a square into an oval pass or an oval into a square pass A deformation can produce any type of bar bullA definite bar into two passes (an intermediate pass followed by a definite pass configures a lsquosequencersquo A sequence only produces a definite bar

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Deformation amp Sequence

bullA definite bar into one intermediate pass or an intermediate bar into one definite pass configures a lsquodeformationrsquo For example a square into an oval pass or an oval into a square pass A deformation can produce any type of bar bullA definite bar into two passes (an intermediate pass followed by a definite pass configures a lsquosequencersquo A sequence only produces a definite bar

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

The roll pass design for any product depends

on the following

bull Starting size and Material Grade

bull Mill layout

bull Mill stand sizes

bull Mill motor power

bull Production Requirement

bull Product size and shape

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Typically a pass design calculation has three

parts

bull Pass design and groove details

bull Pass schedules

bull Power calculation

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Pass Design and Groove Details This calculation

gives the following parameters for each pass

1048707 Roll groove dimensions

1048707 Roll gap

1048707 Filled width in pass

1048707 Filled area

1048707 Area reduction

1048707 Bite angle

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Pass Schedules Pass schedule consists of the

following for each pass

1048707 Bar length

1048707 Rolling speed

1048707 Rolling time

1048707 Idle time

1048707 Loop or tension value between stands

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Power Calculation Power Calculation works out

for each pass

1048707 Bar Temperature

1048707 Rolling load

1048707 Rolling torque

1048707 Rolling power

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Throughout the mill

bullContinuous rolling process -

the long axis of the bar is

brought between the rolls

and is rolled into a shape

with equal axes then this

shape is rolled into a

different shape with different

axes and so on The

reduction must be applied

after a 90-degree rotation of

the bar at each stand

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Throughout the mill bullTraditional mills only use horizontal stands The ovals are twisted to bring the long axis between the rolls

bullTo be precise there is one deformation that needs special treatment the square-into-oval It needs rotating the square by 45deg which can be obtained (if we dont want to use twister guides) with a slight axial displacement of one roll in the stand that produces the square

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

The Mills bullStructures and schematizations bullContinuous bar mill (CBM) structure consists of a number of independent stands Independent means that each stand has its own motor (and kinematic chain) whose rotational speed can be freely altered If you dont want the bar to be twisted you use the HV mill configuration (with definite passes in vertical stands) bullFrom the roll pass design point of view a CBM can be schematized as a succession of passes centered on the z-axis (when xy is the plane containing the roll axes)

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Billet Size Area

Finished

Size

Area of

finished bar

Co-efficient

of

elongation

No of

passes

150 22500 12 1131429 1988636 2078453

150 22500 16 2011429 1118608 1852503

150 22500 20 3142857 7159091 1677243

150 22500 22 3802857 5916604 1602385

150 22500 25 4910714 4581818 1501982

150 22500 28 616 3652597 1412972

150 22500 32 8045714 279652 1308094

150 22500 36 1018286 2209596 1215586

150 22500 40 1257143 1789773 1132834

Number of passes required

No of Passes= log of co-eff of elongationlog(129)

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullFirst Law

The purpose of the rolling process is to start from a

relatively short bar with a large section area aiming to

obtain a very long product with a small section area

Then the first law to remember is that the volume (or

the weight) is a constant from a 12-ton billet you

will obtain a 12-ton coil Cross sectional area times

bar length is a constant (this is not strictly true for

CBMs some weight will be lost with scale and crop

ends but we can afford to neglect that loss)

Laws of Rolling

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullSecond Law

There is another important law to remember the flow is

also a constant Say that the exit bar from stand 1

has cross sectional area = 3467 sq mm and the

finished round has cross-sectional area = 113 sq mm

(hot bar dimensions) If the finished stand delivers at a

speed of 12 mps then stand 1 must run at 039 mps

03 x 3467 = 12 x 113 In this case the constant is

about 1050 ie if you know the areas you can

immediately calculate the exit speeds And you have

no problems in setting the speed at each stand as

each stand has its own independent motor

Laws of Rolling

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullWhen rolling we can identify one action and two

reactions

If we focus on a horizontal stand of a continuous mill for

rounds we see

- that the rolls apply a reduction (vertically)

- that this reduction produces a wanted elongation

- that reduction produces a spread (sideways)

Action amp Reactions

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullWhen the steel is compressed in the rolls it will obviously move in the direction of least resistance so usually there is not only longitudinal flow but also some lateral flow This is called lsquoSpreadrdquo it is generally accepted that beyond a ratio widthheight = 5 spread becomes negligible

Δh ndash the absolute draught in the pass ho ndash stock thickness before the pass R ndash roll radius fndash coefficient of friction

The coefficient of Spread Beta is the ratio between exit and entry width and is normally gt 1

Spread

Δb=115 X Δh

2ho (radicR X Δh- Δh )

2f

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

1Reduction (with a coefficient of reduction Gamma)

2Elongation (with a coefficient of elongation

Lambda)

bullGamma (defined as ratio between exit and entry height) is

always lt 1 If we reduce a 100x10 flat to 8 mm (a 20

reduction) Gamma=08

bullLambda (defined as ratio between exit and entry length but

more often as ratio between entry and exit section area) is

always gt 1 In the example above (100x10 reduced to 100x8)

Lambda = 1000800 = 125 Note that Beta = 1 (100100 =1)

Reduction and Elongation

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

The Dimensions to be taken for aligning

rolls and adjusting roll pass for Box

groove amp flat oval groove

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

The Square Pass

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullSquare Dimensions A 90deg square with sides and corner radius r has area

A=s^2-086r^2 (1)

and actual reduced diagonal

d=sradic (2)-083r (2)

Note Square grooves generally have facing angles alpha = 90deg only for larger squares Generally facing angle alpha is taken as 90deg for s gt 45 mm 91deg down to 25 mm and 92deg for s lt= 25 mm

In these cases the actual reduced diagonal has length

d=ssin(alpha2)+2r(1-1(sin(alpha2)) (3)

Important Formulae

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Oval Pass

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullOval Radius

bullAn oval pass is made of two circular arcs with facing

concavities Three dimensions are considered referring

either to pass or to bar

i b1t = theoretical oval width (pass not physically

measurable)

ii b1r = actual oval width (bar physically measurable)

iii maxw = maximum oval width (pass physically

measurable)

Important Formulae

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

bullOval Radius To identify oval height we only need two dimensions

i h1t = theoretical oval height (pass physically measurable) ii h1r = actual oval height (bar physically measurable)

To draw the oval groove we need to know its radius R The formula is

R=(b1t^2+h1t^2)(4h1t) (4)

Now when gap=0 we have b1t=maxw This means that if the oval is identified as maxw x h1t we can put H=h1t-gap and calculate

R=(maxw^2+H^2)(4H) (5)

Important Formulae

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Shape rolling of initial billet with

initial cross section 100x100 mm2

to 30x30 mm2 consisting of

sequential passes of square-oval-

square passes

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Shape rolling of Cylindrical Bar

Thank You

Letrsquos share and make

knowledge free

Thank You

Letrsquos share and make

knowledge free