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ContentsInitial Problem Statement 2 Narrative 3-8 Solutions 9-14 Appendices 15-16

Bending MetalHow can metal sheets and pipes be bent so that their strength and performance are preserved?

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How can metal sheets and pipes be bent so that their strength and performance are preserved?

Many manufactured and engineered products

make use of metal sheets or pipes that are

bent to form part of a structure or working

mechanism. When a metal sheet is bent the

inner side is compressed and the outer side is

stretched. The amount of stretch depends upon

how the bend is made. For a sharp crease in a

metal sheet, the stretching can be extreme and

can tear the outer side of the metal, weakening

the product. Bending a pipe in such a manner

not only weakens it structurally, it also constricts

the pipe making it less effective for carrying

fluids.

Bending MetalInitial Problem Statement

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Narrative Introduction

MultimediaThe animation Bending Metal Animation is available to introduce this example.

Activity 1You have to manufacture a metal bracket that can be used to hold together two pieces of wood in the following way. What would your design look like if you wanted to avoid a sharp crease in the metal?

Figure 1.

HintYou need to have metal that lays flat against both pieces of wood.

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2. Bending the metalTo avoid sharp creases in metal plates and pipes they are usually bent around the arc of a circle to reduce the stretching on the outer side.

Figure 2.

The radius of the circle around which the bend is made is called the bend radius.

Figure 3.

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Activity 2Consider the design below which has flat sides of length a and b. Would the length of metal you require to it be

(A) a + b ?(B) more than (a + b) ?(C) less than (a + b) ?

Figure 4.

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3. Bending allowance To account for the metal curving around the bend you need to take the length of the flat sections and add a bend allowance, s, so that the actual length is given by: l a b s= + +

The problem now is how to calculate the bend allowance and hence how much metal is required to make the fabrication.

DiscussionLooking again at a schematic of our simple design, the material on the inside is compressed at the bend so is a little shorter than the original piece of metal before bending. The metal on the outside is stretched so that it is a little longer than the original piece of metal before bending. The dotted line in the diagram is called the neutral line. What can you say about the length of the metal on the neutral line inside the metal bar?

Figure 5.

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Activity 3In the design shown below the metal thickness, t, is 4 mm and the lengths a and b are 400 mm and 600 mm respectively. These sides form an internal angle of 120°. The bend radius is 25 mm. Assuming the neutral line is along the centre-line of the metal what length of metal should we start with?

Figure 6.

Hint HintRemember the dotted neutral line neither stretches nor compresses. The length of this line is therefore the same in the flat and bent configurations.

What is the radius of curvature on the neutral line?

HintThe 120° angle is not the angle through which the metal is bent. Identify this angle and calculate its value. What do you notice about its value?

Activity 4Write down a general expression for the bending allowance, s, of a piece of metal of thickness t, bent through an angle θ with a bend radius r.

HintRemember where the neutral line is.

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4. More complex designsActivity 5Work out the total length of metal for the following design.

Figure 7.

MultimediaThe resource Bending Metal Interactive is available to show different configurations. See appendix 1.

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Solution Introduction

Activity 1 solutionOne way would be to use two pieces of metal and then weld them at the appropriate angle. You would have to be sure that the weld was strong for this to be a reliable solution.

Another method would be to make it out of a single piece of metal but instead of making a sharp crease, make the bend around cylinder so that the curve is gentler.

Figure 8.

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2. Bending the metalActivity 2 solutionThe length of metal is greater than (a + b) as the metal has to curve around the bend.

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3. Bending allowanceFirst construct a fully labelled diagram.

Figure 9.

The first thing to do is work out the bend allowance. This is the length of the curved arc of metal. To do this you need to work out the length r and the angle θ. The length r is the radius of the neutral line. The length of the arc on this circle gives you the bend allowance.

The angle θ can tell you what fraction of a complete circle the arc makes. You can use this to work out the arc length which, by definition is the bend allowance, s, by multiplying this fraction by the circumference of a circle of radius r:

s c= ×θ

360�

where c is the circumference of the circle of radius r.

The circumference of a circle is given by:

c r= 2π

Substituting into the expression for bend allowance, s

s c r r= × = × =θ θ θ

3602

360 180� � �ππ

The value of θ is such that the sum of θ and the internal angle between the flat sides equals 180°.

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Using the values of r = 27 and θ = 60 in the expression for the bend allowance, s, gives

s r= =×

× = ( )π πθ180

60180

27 28 3. ( )mm 1 d.p

So that the total length of metal, l, is given by l a b s= + +

= + + = ( )400 600 28 3 1028 3. . mm

The angle θ through which the metal is bent is calculated by considering the angles in the diagram.

O

A

B

C 120°

θ

D

Figure 10.

Method 1: triangles

The angle θ is given by ∠AOC. ∠ = − =BDC 180 120 60� � �

Looking at triangle BCD: ∠ +∠ + =BDC DCB 90 180� � so

∠ = − −∠

= − −

=

DCB BDC180 90180 90 6030

� �

� � �

�

Now note that ∠ = ∠DCB ACO=30�

Looking at triangle AOC: ∠ +∠ + =AOC ACO 90 180� � but ∠ =AOC θ , and you have just found an expression for ∠ACO so

θ

θ

θ

+∠ ∠ ∠

+∠ ∠

+

OAD+ ADB+ DBO=360OAD+ + DBO=360

+ + =3

�

� �

� � �

12090 120 90 660

360

�

� � � �

�

θ = − − −

=

90 120 9060

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Method 2: quadrilaterals

Consider the quadrilateral OADB. The angles of a quadrilateral sum to 360°

θθ

θθ

+∠ ∠ ∠+∠ ∠

+=

OAD+ ADB+ DBO=360OAD+ + DBO=360

+ + =36036

12090 120 90

00− − −=

90 120 9060

The value of θ is such that the sum of θ and the internal angle between the flat sides equals 180°.

Using the values of r = 27 and θ = 60 in the expression for the bend allowance, s, gives

s r= =×

× = ( )π πθ180

60180

27 28 3. ( )mm 1 d.p

So that the total length of metal, l, is given by

l a b s= + +

= + + = ( )400 600 28 3 1028 3. . mm

Activity 4 solutionThe bend allowance is made by considering the arc length of the neutral line which runs down the centre of the metal. Therefore, the bend radius of the neutral line is given by the bend radius plus half the thickness of the metal:

r r tneutral = +

2The bend allowance, s, is the arc length on the neutral line. This is given by the fraction of the circumference that a bend of angle q makes so that

s r r tneutral= = +

π πθ θ180 180 2� �

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4. More complex designs Activity 5 solutionThis design is an extension of the previous one. The additional lengths are 800 mm from the upper flat and the bend allowance for the upper curve, which need to be added to the value calculated previously. The angle between the left flat and the upper flat is shown as 60°. Using the rule that the arc angle plus the internal angle equal 180° it is seen that the arc angle is 120°. The that the bend allowance on the upper curve is therefore given by

s r= =×

× = ( )π πθ180

120180

52 108 9. ( )mm 1 d.p

Note that the inner bend radius is 50 mm but the neutral line has a radius of this plus half the material thickness; hence an effective radius of 52 mm.

The total length is then

where the 1028.3 was calculated in the previous activity.

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Appendix 1using the interactives

Bending Metal InteractiveThis resource demonstrates the calculation of metal length required to produce a bend sheet.

At the bottom of the screen is a bar that lets you choose whether you want to try finding a variable for a single bend of metal, a double bend of metal or a bar of metal with a random number of bends.

Figure 11.

You will be given the information for each known variable and are required to fill in the missing value into the empty box. You can use the diagram to help you do this, giving your answer to the nearest whole value.

You can check your answer by clicking on the “Check” button in the bottom right hand corner of the screen.

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Appendix 2mathematical coveragePL objectivesUse trigonometry and coordinate geometry to solve engineering problems• Use both degrees and radians and convert between them

Use algebra to solve engineering problems• Solve problems involving ratio and proportion• Solve problems involving area, perimeter and volume