Mathematics and Steel Pressing Mathematics-and-steel-pressing.pdf

7/24/2019 Mathematics and Steel Pressing Mathematics-and-steel-pressing.pdf

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Mathematics and steel roll pressing

Part 1Making 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


(For teacher to present to students as considered appropriatecould 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 steelrolling 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.
http://twwoods.com.au/gallery.phphttp://twwoods.com.au/gallery.phphttp://twwoods.com.au/gallery.phphttp://twwoods.com.au/gallery.php


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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 clipswill 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 alongthe 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.
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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)


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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 cant 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
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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.)

It is most important to have the steel plate cut very accurately, (within 1 mm), to the correctlength and width before it is pressed. The pressing process cant be adjusted to correct any errors

in the dimensions of the plateif 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= 2r= 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 5655mm 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.

1500 mm1800 mm

60 mm


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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 (90030) mm or870 mm. This was obtained from the formulaRM=RE T, whereRMis the radius of the

mean running,REis the external radius and Tis 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 radiiworth discussing with

studentsusing 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= 2RM 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+ 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


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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 cant 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.


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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

AB

C
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Radius of mean running = bend radius + thickness

= 20 + 5

= 25 mm

Note that the bend angle is 900, which is of a revolution.

Bend allowance = of circumference of mean running circle

= 2 25

= 39 mm (correct to the nearest mm)

Step 3 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 = 500mm).

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 400bend upwards with a bend radius of 1200mm, followedby 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 radius + thickness

= 1200 + 10

= 1210 mm

Bend allowance =of circumference of mean running circle

= 2 1210

= 845 mm (correct to the nearest mm)

Step 2 Calculate length required


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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 2Making 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


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.


Background information


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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= 120

2+ 50

2

= 14400 + 2500

= 16900

s= = 130

The slant height of the cone is 130 mm, so we will need todraw a circle of radius 130 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 dont need to use our calculator

120 mm

50 mm

s

The arc of this

sector is rolled to

make the

circumference of

the base of the

cone
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to find this value.)

The circumference of the base of the

cone is C= 2 50 which is

equal to 100 mm.

What fraction of the required circle

is the circumference of the base ofthe cone?

or

.

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 frustumthey 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 radiiyou 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.

130 mm

1380

The radii of this

sector are joined

together along

the slope of the

cone.


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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 pieceshas 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 theradii 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)
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The slant height is 381 mm

Step 2 Calculate the radii of the annulus

Method 1 (Use similar triangles)

We note that OPB/// AQB.


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We use the formula

whereRsectoris the radius of the larger of the concentric

circles,Ris the larger radius of the cone, ris the smaller radius of the cone and lis 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 635381 = 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= .


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In AQB,

We dont need to work out the value of .

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 635381 = 254 mm.

The concentric circles will have radii of 635 mm and 254 mm.

Step 3 Calculate the sector angle

We use the formula: Angle of sector =

3600.

Sector angle = = 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.
http://twwoods.com.au/files/1266202364-image_2.jpg


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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 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 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 5164906 ).

Step 3 Calculate the sector angle

SA=

3600where SAis the sector angle.

SA=

= 400

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


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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 Calculatorwhich 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.)


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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 100bend 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.


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7. A mathematician analysed the problem of calculating the bend allowance and produced the

formulaBA= (0.0175R+ 0.0087 T ) .

In this formulaBAis the bend allowance,Ris the bend radius, Tis the thickness of the steel

plate, and 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.

50 mm

1000 mm

200 mm

1200 mm

900bend at

300 mm radius

500

bend at

100 mm radius

900 mm

900bend at

100 mm radius


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(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

= 2 (radius to the middle of the path ) width of path thickness

Use the old-timers 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-timers 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.

Loubies answer has been voted as the best answer to Johns question.

Carefullycheck Loubies answer.

Would you vote it as the best?


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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 activityBefore 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)


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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 fitsimilar to what happens in industrycentre of concentric circles is

not on the actual steel plate that is going to be marked and cutcould be metres away for a cone

with only very slight taperleads into use of sagitta or versines will do as a separateresource)

2. Find the dimensions of the shape that has to be rolled to make:

a) this lamp shade

b) this flower pot

c) this waste bin

160 mm

100 mm

95 mm

200 mm

400 mm

180 mm
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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) a conical piece from 50 mm plate with height 800 mm and diameters of 500 mm and 350

mm

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

2. (a)

2

2 TRL

E (b)

2

2 TRL

I (c)

22

EI RR

L or EI

RRL

4.(a) 1148 mm (b) 1970 mm (c)1886 mm

180mm

170 mm

250 mm
http://www.officeking.com/image/cache/data/VOW2012/KF12011-500x500.jpghttp://www.officeking.com/image/cache/data/VOW2012/KF12011-500x500.jpghttp://www.officeking.com/image/cache/data/VOW2012/KF12011-500x500.jpghttp://www.officeking.com/image/cache/data/VOW2012/KF12011-500x500.jpg


26/26

5. 2711 mm

6. 4553 mm

7. (b) The formula to give the exact value of the bend allowance is TRBA 5.02360

or

TRBA360360

2. Now 017453292.0

360

2

and 008726646.0

360

, hence the

mathematicians 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 225

0 (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

Mathematics and Steel Pressing Mathematics-and-steel-pressing.pdf

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