roll pass design in continuous bar mills
DESCRIPTION
taken from ANSAR HUSSAIN RIZVI Manager (Production)TRANSCRIPT
![Page 1: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/1.jpg)
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
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![Page 2: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/2.jpg)
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
![Page 3: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/3.jpg)
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
![Page 4: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/4.jpg)
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
![Page 5: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/5.jpg)
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
![Page 6: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/6.jpg)
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
![Page 7: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/7.jpg)
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
![Page 8: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/8.jpg)
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
![Page 9: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/9.jpg)
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
![Page 10: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/10.jpg)
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
![Page 11: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/11.jpg)
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
![Page 12: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/12.jpg)
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
![Page 13: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/13.jpg)
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
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![Page 14: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/14.jpg)
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
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![Page 15: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/15.jpg)
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
![Page 16: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/16.jpg)
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
![Page 17: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/17.jpg)
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
![Page 18: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/18.jpg)
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
![Page 19: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/19.jpg)
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
![Page 20: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/20.jpg)
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
![Page 21: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/21.jpg)
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
![Page 22: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/22.jpg)
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
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Letrsquos share and make
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![Page 23: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/23.jpg)
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
![Page 24: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/24.jpg)
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
![Page 25: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/25.jpg)
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
![Page 26: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/26.jpg)
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
![Page 27: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/27.jpg)
Shape rolling of Cylindrical Bar
Thank You
Letrsquos share and make
knowledge free
![Page 28: Roll pass design in continuous bar mills](https://reader034.vdocuments.net/reader034/viewer/2022042506/547a0f0db379593a2b8b48fc/html5/thumbnails/28.jpg)
Thank You
Letrsquos share and make
knowledge free