Mathematics and steel roll pressing
Part 1 – Making cylinders and bends by roll pressing steel
Outcomes for Part 1(Old syllabus):
Maths: NS4.3, NS5.2.1, PAS4.1, PAS4.4, MS4.1, MS4.2, WMS4.2, WMS4.3, WMS4.4
ICT: enter formulae into spreadsheets, create and format spreadsheets
Outcomes for Part 1(New syllabus):
MA4-1WM communicates and connects mathematical ideas using appropriate terminology, diagrams and
symbols
MA4-2WM applies appropriate mathematical techniques to solve problems
MA4-3WM recognises and explains mathematical relationships using reasoning
MA4-5NA operates with fractions, decimals and percentages
MA4-12MG calculates the perimeters of plane shapes and the circumferences of circles
ICT: enter formulae into spreadsheets, create and format spreadsheets
(For teacher to present to students as considered appropriate – could be by teaching using
selected material from this package, or by allowing students to work through the background
material themselves, or could be adapted as an “Assessment for learning” activity- see ‘steel
rolling a4l.doc’)
Background information
Rolled steel is used extensively in industry.
Steel sheet, (a few mm thick), or steel plate, (6 mm or more in thickness), is pressed into
cylinders and cones, and steel can be pressed around specific arcs to form bends or angles. These
are pressed to great accuracy and are then assembled to make bulk storage containers, ductwork,
piers for oil rigs, panels for the hulls of ships, etc.
TW Woods Construction is a heavy engineering and steel fabrication company based at Tomago,
NSW. The gallery of photographs, (view photos here), gives a glimpse of the scope and size of
the applications.
The steel plates that are used in the production of these items range in thickness from 6 mm to 70
mm.
TW Woods has several hydraulic roll presses that are used to bend the steel plate into the
required shape.
The following videos show steel being pressed to form cylinders, angles and cones.
Cylinder being pressed
Angle being pressed
Cone being pressed
Note these are YouTube clips – will need to have download copies saved
We hope to get equivalent clips from TW Woods.
Typically the finished products are accurate to within 1or 2 mm of specifications.
During the pressing process the steel is deformed in such a way that matter is ‘squished’ along
the length from the inner surface of the steel plate to the outer surface. i.e. the inner half of the
steel is compressed as it is forced around the roller while the outer half is stretched. The section
of the plate that is exactly in the middle, (with regard to the thickness), is not affected.
The length of the middle section of the steel plate is called the ‘mean running’.
What are the implications of this?
We know that the label that wraps around a tin of peaches is very thin. The label is actually
printed on a rectangular sheet of paper which is then rolled to be glued to the tin. The length of
the rectangular piece of paper is equal to the circumference of the cylindrical can.
The length of the label is approximately 230 mm.
The diameter of the tin is approximately 73 mm. Confirm these results using C =  d.
For a cylinder made from thin material such as paper, cardboard, or sheet metal, the internal and
external radii of the cylinder are virtually identical. (There is a difference, but it is so small as to
be negligible).
Example 1 Making a cylinder in sheet metal (1 or 2 mm thick)
Flues take the waste gases and smoke from a gas or wood heater and vent them outside the
house.
A flue is made from cylindrical sections which are joined together. Sections are typically 900
mm or 1000 mm in length. The diameter may be from 100 mm to 305 mm (or more) depending
on the size of the heater. The sections are rolled from sheets of stainless steel, galvanized steel,
or copper depending mainly on the appearance required.
Similar cylindrical components may be used in ducting for air conditioning.
The picture shows part of the flue for a home fire place. It is
made from rolled stainless steel. It has a diameter of 230 mm
and is 900 mm in length.
Find the dimensions of the rectangle of stainless steel sheet
from which each section is made. (Allow an additional 5 mm
on the length to allow the edges of the rolled cylinder to be
overlapped and joined together).
Using C =  d, we find the circumference of the cylinder to be  × 230 which is 723 mm (to the
nearest mm).
Now we need to add on the 5 mm joining allowance.
The stainless steel sheet would need to be cut to a length of 728 mm and a width of 900 mm.
However, for a cylinder made from steel which is 60 mm thick, or even 6 mm thick, the
difference between the internal and external radii can’t be ignored.
When the thickness is 6 mm or more it is called plate steel rather than sheet steel.
Example 2 Pressing a cylinder from steel plate
Suppose that we have to make a cylinder from 60 mm steel
plate such that the cylinder has a height of 1500 mm and
external radius of 900 mm. (Note that in the manufacturing
industry lengths are expressed in mm.)
60 mm
1500 mm
1800 mm
It is most important to have the steel plate cut very accurately, (within 1 mm), to the correct
length and width before it is pressed. The pressing process can’t be adjusted to correct any errors
in the dimensions of the plate – if we press an incorrectly sized plate we will get an incorrectly
sized cylinder.
We know that the steel plate will have to be 60 mm thick and 1500 mm wide, but how long must
it be?
It appears that the length of the plate will have to be equal to the outside circumference of the
cylinder C = 2 r = 2 ×  × 900 = 5655 mm (correct to the nearest mm).
If we act on this calculation and cut a sheet of 60 mm thick steel into a rectangle of length 5655
mm by 1500 mm, then roll it, we will get a cylinder of the correct height but the radius will be
incorrect.
Why? In the rolling process the outside material is stretched as it is rolled, which means that the
outer circumference will be more than 5655 mm, which then means that the external radius will
be greater than 900 mm.
How do we solve the problem? We must work with a length measurement that does not change
during the rolling process, i.e. we must work with the mean running.
Step 1 Calculate the radius of the mean running
The radius of the circle produced with the mean running as its circumference is (900– 30) mm or
870 mm. This was obtained from the formula RM = RE − ½ × T, where RM is the radius of the
mean running, RE is the external radius and T is the thickness.
This was calculated by subtracting half of the thickness from the external radius.
Which explanation is better for the target students?
(Note that people in trades usually work with diameters rather than radii – worth discussing with
students – using calipers etc you measure diameter, not radius).
Step 2 Calculate the circumference of the mean running, which is the required length to cut
Using C = 2 RM we obtain C = 2 ×  × 870 = 5466 mm (correct to the nearest mm).
Now we know that the 60 mm thick steel plate has to be cut to a length of 5466 mm and a width
of 1500 mm.
Example 3 Pressing a cylinder from steel plate
We are required to manufacture a cylinder from plate steel.
The dimensions of the cylinder are:
height 800 mm
inside diameter 1200 mm
wall thickness 70 mm
Find the dimensions of the steel plate to be used.
Step 1
The mean radius is 635 mm.
Note that when working with inside units, RM = RI + ½ × T. (i.e. the radius of the mean running
is the interior radius plus half the thickness).
Step 2
Length required
= circumference of mean running
= 2 ×  × 635
= 3990 mm
The steel required is 3990 mm long, 800 mm wide and 70 mm thick.
A video of entire process (order specs to cutting to pressing to welding) at TW Woods will be
included
Example 4 Pressing a bend (angle) from steel plate
Thin materials like paper or cardboard can be folded along a line to form any required angle.
Steel plate can’t be folded along a line to form an angle. It has to be pressed and will produce the
required bend around a circular arc. This is shown in the following picture.
This pic from Jorgensen Metal Rolling site, do we acknowledge copyright. Would prefer to get
something similar from TW Woods.
We need to be familiar with some specific terminology, as shown in the following diagram.
A
B
C
A combination of these two videos might be useful to present and explain the terminology,and to
explain why the bend angle is equal to the angle at the centre of the circular arc.
Terminology used in bending
Bend angle and angle at centre
Suppose that we have to make a right angle bend in 10 mm steel plate
such that the finished product is 800 mm wide, the mold line
measurements are 200 mm on one side and 300 mm on the other, and the
bend radius is 20 mm.
What are the dimensions of the steel plate that has to be pressed?
Clearly the steel plate will be 10 mm thick and 800 mm wide. The problem is in calculating the
length of plate required, because there will be the same type of compression on the inside and
stretching on the outside of the arc formed by pressing as there was when rolling a cylinder.
Notice the term ‘bend allowance’ shown on the diagram. It is the length of the arc of the mean
running.
Step 1
Calculate the distance from the edge of the steel plate to each bend line
Distance to bend line = mold line measurement – ( bend radius + thickness)
= mold line measurement – (AB + BC) as shown on the diagram
= 200 – (20 + 10)
and
300 – (20 + 10)
= 170 mm
and
270 mm
Step 2
Calculate the bend allowance
Radius of mean running
= bend radius + ½ × thickness
= 20 + 5
= 25 mm
Note that the bend angle is 900, which is ¼ of a revolution.
Bend allowance
Step 3
= ¼ of circumference of mean running circle
= ¼ × 2 ×  × 25
= 39 mm (correct to the nearest mm)
Calculate length required
Length required
= Sum of distances to bend lines + bend allowance
= 170 + 270 + 39
= 479 mm
Note that this is significantly less than the sum of the two mold line distances (200 + 300 = 500
mm).
The 10 mm thick steel plate has to be cut to a length of 479 mm and width of 800 mm.
Example 5 Pressing a bend in steel plate
We have to make a bend in 20 mm steel plate as part of a chute. The
chute is to be 700 mm wide. It consists of a 1300 mm long flat section,
leading to a 400 bend upwards with a bend radius of 1200mm, followed
by a straight section of length 500 mm.
What are the dimensions of the steel plate that has to be pressed?
Step 1
Calculate the bend allowance
Radius of mean running
Bend allowance
= bend radius + ½ × thickness
= 1200 + 10
= 1210 mm
= of circumference of mean running circle
= × 2 ×  × 1210
= 845 mm (correct to the nearest mm)
Step 2
Calculate length required
Length required
= Sum of distances to bend lines + bend allowance
= 1300 + 500 + 845
= 2645 mm
The 20 mm thick steel plate has to be cut to a length of 2645 mm and width of 700 mm.
Students would now be given Activity 1(or the parts of it suited to their ability).
Part 2 – Making conical pieces by pressing steel
Outcomes for Part 2 (Old syllabus):
Maths: NS4.3, NS5.2.1, PAS4.1, PAS4.4, PAS5.2.2, MS4.1, MS4.2, MS5.2.2, SGS5.2.2,
WMS4.2, WMS4.3, WMS4.4
ICT: enter formulae into spreadsheets, create and format spreadsheets
Outcomes for Part 2 (New syllabus):
MA4-3WM recognises and explains mathematical relationships using reasoning
MA5.1-1WM uses appropriate terminology, diagrams and symbols in mathematical contexts
MA5.1-2WM selects and uses appropriate strategies to solve problems
MA5.2-2WM interprets mathematical or real-life situations, systematically applying appropriate strategies
to solve problems
MA4-5NA operates with fractions, decimals and percentages
MA4-10NA uses algebraic techniques to solve simple linear and quadratic equations
MA4-12MG calculates the perimeters of plane shapes and the circumferences of circles
MA4-16MG applies Pythagoras’ theorem to calculate side lengths in right-angled triangles, and solves
related problems
MA5.1-10MG applies trigonometry, given diagrams, to solve problems, including problems involving
angles of elevation and depression
MA5.2-13MG applies trigonometry to solve problems, including problems involving bearings
MA5.1-11MG describes and applies the properties of similar figures and scale drawings
Depends on approach used.
ICT: enter formulae into spreadsheets, create and format spreadsheets
Background information
The following video shows how a cone can be rolled from a sector of a circle.
video of making a paper cone (need to record the equivalent of first 3 minutes of video)
Example 6 Making a cone from a sheet of cardboard
We have to make a cone which has a height of 120 mm and a diameter of
100 mm.
What are the dimensions of the circular sector that must be cut out from
the cardboard to make this cone?
Step 1
Find the radius of the required circle
The radius of the circle is the ‘slant height’ of the cone. The video shows that when the circular
sector is rolled to make the cone the radii on each end of the sector are taped together along the
sloping side of the cone.
We find the slant height of the cone using Pythagoras theorem.
s2= 1202 + 502
= 14400 + 2500
= 16900
120 mm
s
s =√
= 130
The slant height of the cone is 130 mm, so we will need to
draw a circle of radius 130 mm.
50 mm
Note: Weaker students could just be given the formula
)
(
)
√(
Top students could be asked to explain why the formula works.
Step 2
Find the size of the angle at the centre of the newly drawn circle
The newly-drawn circle of radius
130 mm has a circumference C = 2 ×
 × 130 which is equal to 260 mm.
(We don’t need to use our calculator
The arc of this
sector is rolled to
make the
circumference of
the base of the
cone
to find this value.)
The circumference of the base of the
cone is C = 2 ×  × 50 which is
equal to 100 mm.
1380
130 mm
What fraction of the required circle
is the circumference of the base of
the cone?
or .
The radii of this
sector are joined
together along
the slope of the
cone.
This tells us that we need of the
constructed circle.
How do we get of a circle?
We measure an angle at the centre of
the circle which is
, which
is
, then cut along the two radii.
We use the sector with the
angle to roll the cone.
Note: Weaker students could just be given the formula
Angle of sector
Top students could be asked to explain why the formula works.
Example 7 Making a frustum of a cone from a sheet of copper
A frustum is the name given to the solid formed when a pyramid or cone has a top section
removed. In industry the manufacturers would not use the term frustum – they just call them
cone sections or conical pieces.
Do the following activity:
On a sheet of paper draw two concentric circles (i.e. with the same centre). Make the radii
100mm and 60 mm.
1. Draw two radii – you can choose them to be at any angle you want. Use a thicker red pen
or marker to draw along the two radii.
Use a thicker blue pen or marker to draw along the minor arc of each circle.
Use a thicker green pen or marker to draw along the major arc of each circle.
2. Carefully cut out the annulus (the part between the two concentric circles). Keep the
annulus, put aside the piece of paper which was the small inner circle, and discard the
piece of paper which was the outside part of the big outer circle.
3. Now cut along the radius lines that are on the annulus.
4. The annulus has now been cut into two strips of paper. Each is a sector of the annulus.
Roll each strip into a section of a cone.
5. Reassemble the 3 pieces of paper and use the colours of the different sides and arcs to
confirm that the arcs of the concentric circles become the circumferences of the top and
bottom circles of the conical frustum and that the slant height of the frustum is formed by
the radius section of the original sector.
We have to make a conical section from galvanized steel sheet to join
together two pieces of ducting which are 350 mm apart and have their
ends directly aligned with each other (centre to centre). One of the pieces
has diameter 500 mm while the other piece has diameter 200 mm.
The picture shows some additional rolling on the top and bottom. We
will ignore this.
Find the dimensions of the shape which will have to be rolled to form the
required section.
We know that the shape that has to be rolled is a sector of an annulus. We need to work out the
radii of the concentric circles and the sector angle.
Step 1
Calculate the slant height of the conical piece
Note that
√
i.e.
√
(
(
)
)
(to the nearest mm)
The slant height is 381 mm
Step 2
Calculate the radii of the annulus
Method 1
(Use similar triangles)
We note that OPB /// AQB.
We use the formula
where Rsector is the radius of the larger of the concentric
circles, R is the larger radius of the cone, r is the smaller radius of the cone and l is the slant
height of the cone.
(to the nearest mm)
We now need to find the radius of the smaller of the concentric circles.
This radius will be equal to
.
i.e. the radius of the smaller circle is 635 – 381 = 254 mm.
The concentric circles will have radii of 635 mm and 254 mm.
Method 2
Use trigonometry
Note that POB = QAB (corresponding angles in parallel lines, OP // AQ.
Let POB = QAB = θ.
In AQB,
We don’t need to work out the value of θ.
In OPB,
We now need to find the radius of the smaller of the concentric circles.
This radius will be equal to
.
i.e. the radius of the smaller circle is 635 – 381 = 254 mm.
The concentric circles will have radii of 635 mm and 254 mm.
Step 3
Calculate the sector angle
We use the formula:
Sector angle =
Angle of sector =
× 3600.
= 1540
Example 8 Making a section of a cone from a steel plate
The picture shows the final stages of the rolling of a conical section from 60 mm steel plate.
The cone section will be 900 mm in height and will have external diameters of 1200 mm and
1000 mm.
Find the dimensions of the annular sector that will be rolled to form the cone section.
Remember that, when rolling thick steel, we must use the mean running to calculate the length,
but not to calculate the slant height.
This means that we have to work out the radius of the mean running at both the top and bottom
of the cone section, then use the mean running radii in the formulae.
Step 1
Calculate the slant height of the cone and the radii of the mean running
The slant height is
√
(
)
(to the nearest mm)
At the larger end of the cone piece, the outside radius is 600 mm and the thickness is 60 mm, so
R = 570, obtained from 600 − ½ × 60.
At the smaller end of the cone piece, the outside radius is 500 mm and the thickness is 60 mm, so
r = 470, obtained from 500 − ½ × 60.
Step 2
Calculate the radii of the concentric circles
(to the nearest mm)
The concentric circles will have radii of 5164 mm and 4258 mm (from 5164 – 906 ).
Step 3
Calculate the sector angle
SA =
× 3600 where SA is the sector angle.
= 400
SA =
Students would now be given Activity 2 (or the parts of it suited to their ability).
Activity 1
1. Find the length of steel plate required to make each of the following cylinders, answer to the
nearest mm :
a)
wall thickness 50 mm, height 1300 mm, outside radius 600 mm
b)
wall thickness 30 mm, height 650 mm, inside radius 500 mm
c)
wall thickness 40 mm, height 1800 mm, external diameter 1400 mm
d)
height 950 mm, outside radius 670 mm, inside radius 620 mm
e)
wall thickness 30 mm, height 700 mm, outside radius 850 mm
f)
wall thickness 60 mm, height 1650 mm, inside radius 750 mm
g)
wall thickness 20 mm, height 1200 mm, external diameter 1000 mm
h)
height 950 mm, outside radius 1600 mm, inside radius 1540 mm
2. Write down a formula which will calculate the length of steel plate required to make a cylinder
when given:
a)
the outside radius and the wall thickness
b)
the inside radius and the wall thickness
c)
the inside radius and the outside radius
Check that your formula is correct by using Question 1 and its answers as test data.
3. Use a spreadsheet to make an ‘Instant Length Calculator’ which displays the correct length to
cut the steel plate which is to be rolled into a specified cylinder.
There should be three independent sections of the spreadsheet to calculate length required from:
a)
outside radius and wall thickness
b)
inside radius and wall thickness
c)
inside radius and outside radius
Check that your spreadsheet is correct by using Question 1 and its answers as test data.
4. Find the length of steel plate required to make each of the following bends:
a)
plate thickness 50 mm, distance of 400 mm to each of the bend lines, bend radius 260
mm, and the plate is to be bent at an angle of 700.
b)
plate thickness 10 mm, bend line distances of 1000 mm and 800 mm, bend radius 60 mm,
and the plate is to be bent back on itself so that the two faces form an angle of 300. (i.e. the bend
angle is 1500.)
c)
plate thickness 30 mm, mold line distances of 1200 mm and 800 mm, bend radius 180
mm, and the plate is to be bent at 900.
5. We have to make a bend in 20 mm steel plate as part of an access ramp at a golf course to
allow golfers to drive their buggies into a shed. The ramp is to be 1500 mm wide. The ramp
consists of a 500 mm long flat section leading to a 100 bend with a bend radius of 1200mm,
followed by a flat section of 2000mm. Draw a diagram to illustrate this information, then find the
dimensions of the steel plate required.
6. Use the dimensions given on the picture to find the length of steel plate required to make one
of these structural supports.
900 bend at
100 mm radius
200 mm
50 mm
900 bend at
900 mm
300 mm radius
1000 mm
500 bend at
1200 mm
100 mm radius
7. A mathematician analysed the problem of calculating the bend allowance and produced the
formula BA = (0.0175 R + 0.0087 T ) θ.
In this formula BA is the bend allowance, R is the bend radius, T is the thickness of the steel
plate, and θ is the bend angle.
a)
Confirm that the formula is correct by using Questions 4, 5 and 6 and their answers.
(You may find slight differences in the answers.)
b)
Analyse the problem of finding the bend allowance and explain why the formula works.
(i.e. explain the mathematical basis to the formula, and discuss why there are slight differences
between the answers from the formula and your answers to questions 4, 5, 6.)
8. This question is not about steel rolling, but you should see the connection.
A concretor has to work out how much concrete (in cubic metres) to order to make a footpath to
go around the circumference of a circular pond. The pond has a radius of 10 m. The path is to be
1.2 m wide and 100 mm thick.
(i) Work out the volume required, correct to 1 decimal place.
Expect students to do a difference of the volumes of two cylinders, or difference of the areas of
two concentric circles then multiply by the depth.
(ii) ‘Old-time’ concretors would work out their answer using the formula
Vol = circumference of circle in middle of path × width of path × thickness
= 2π × (radius to the middle of the path ) × width of path × thickness
Use the old-timer’s method to work out the volume, and confirm that the method works.
(iii) For the really top students
Explain (i.e. give a mathematical explanation) why the old-timer’s method is equivalent to the
method you used in (i). Requires factorization of difference of two squares.
(iv) Can you see the similarity to calculating lengths for steel rolling?
9. Following on from question 8:
The following screen captures were taken from the UK and Ireland site of YAHOO!
ANSWERS.
Loubie’s answer has been voted as the best answer to John’s question.
Carefully check Loubie’s answer.
Would you vote it as the best?
10. Suppose that a friend asked you for help with working out the dimensions of metal plate
needed to make cylinders. Write a response to your friend in which you explain the process, in
your own words, so that your friend will be able to work out the required measurements.
Activity 2
1. Make the following from paper or cardboard:
a)
a cone of height 80 mm and radius 60 mm
Fun activity – Before rolling the sector into the cone, mark off points on each radius at 10
mm intervals. Use a coloured pen to rule intervals from each mark on the left side radius to the
mark higher up on the right side radius- i.e. produce a set of parallel intervals on the sector.
Observe the effect when the sector is rolled into the cone.
b)
a cone of height 80 mm and radius 120 mm (need A3 size paper)
c)
a section of a cone with height 50 mm and radii of 70 mm and 30 mm
d)
a section of a cone with height 50 mm and radii of 70 mm and 60 mm (needs A3 and
some thinking to make it fit – similar to what happens in industry – centre of concentric circles is
not on the actual steel plate that is going to be marked and cut – could be metres away for a cone
with only very slight taper – leads into use of ‘sagitta’ or ‘versines’ – will do as a separate
resource)
2. Find the dimensions of the shape that has to be rolled to make:
a)
this lamp shade
180 mm
200 mm
400 mm
b)
this flower pot
160 mm
95 mm
100 mm
c)
this waste bin
250 mm
180mm
170 mm
3. Use a spreadsheet to make an ‘Instant Cone Calculator’ which displays the correct radius and
angle to cut the sheet of material which is to be rolled into a specified cone.
Use questions 1a and 1b and your answers to check that your spreadsheet is correct.
4. Use a spreadsheet to make an ‘Instant Conical Section Calculator’ which displays the correct
radii and angle to cut a thin sheet of material which is to be rolled into a specified conical
section.
Use questions 1c and 1d and your answers to check that your spreadsheet is correct.
5. Calculate the dimensions of the shape required to roll the following from steel plate:
a)
a conical piece from 20 mm plate with height 1000 mm and diameters of 900 mm and
700 mm
b)
mm
a conical piece from 50 mm plate with height 800 mm and diameters of 500 mm and 350
6. Modify your spreadsheet from question 4 so that it also includes a calculator for working with
plate steel.
Answers
Activity 1
1. (a) 3613 mm (b) 3236 mm (c) 4273 mm (d) 4053 mm (e) 5246 mm (f) 4901 mm (g) 3707
mm (h) 9865 mm
T
T
 R  RE 


2. (a) L  2  RE   (b) L  2  RI   (c) L  2  I
 or L   RI  RE 
2
2
2




4.(a) 1148 mm (b) 1970 mm (c)1886 mm
5. 2711 mm
6. 4553 mm
7. (b) The formula to give the exact value of the bend allowance is BA 

360
 2 R  0.5T  or
2

 
 2
 0.017453292 and
 0.008726646 , hence the
BA  
R
T  . Now
360
360
360 
 360
mathematician’s formula gives an approximate value.
8. (i) 8.0 m3
Activity 2
1. (a) sector of circle of radius 100 mm, sector angle 2160 (b) sector of circle of radius 144 mm,
sector angle 3000 (c) sector of annulus of radii 112 mm and 58 mm, sector angle 2250 (d) sector
of annulus of radii 357 mm and 306 mm, sector angle 710
2. (a) sector of annulus of radii 415 mm and 187 mm, sector angle 1730 (b) sector of annulus of
radii 251 mm and 157 mm, sector angle 1150 (c) sector of annulus of radii 575 mm and 391 mm,
sector angle 780
5. (a) sector of annulus of radii 4422 mm and 3417 mm, sector angle 360 (b) sector of annulus of
radii 2412 mm and 1608 mm, sector angle 340