Integrating Studies 3 project

Integrating Studies 3—Detail Design Project


Forward and Backward Stepper

Kerrie Noble

200948192

PDE

[email protected]

Forward and Backward Stepper


Contents

Area of Design Page Number

Introduction 1

Initial Concept Sketches 2

Chosen Concept 3

Rough Layout Sketch 4

Embodiment Design 5

Detailed Layout Sketch 6

Force Analysis 7—11

Design for Buckling 12

Fastening and Welding 13

Bearing Selection 14

Mechanica Analysis 15

Manufacturing 16—17

Movement Mechanism Analysis 18

Resistance Mechanism 19—20

Part Sourcing 21

Product Costing and Bill of Materials 22

CAD Rendering and Views 23—26

Manufacturing Drawing Set 27—55


The brief for this project was to design a practical but desirable piece of exercise equipment suitable for

use in the home. The item was to;

Be for a specific exercise type, target user, or sport

Incorporate some form of adjustable mechanical resistance

Be in a neat, compact, packaged form factor suitable for the home

Incorporate an electronic display

This sector of the exercise industry is growing due to the busy schedule and fast pace of life for many

people. This fast pace has resulted in a move from the gym to accessible exercise at home. When the

consumer is thinking about purchasing home exercise items they are considering factors such as;

Exercises they enjoy doing

Their budget

The space they have within their home for the equipment

The legitimacy of the exercise and its results

This means important factors to consider when designing a piece of exercise equipment are;

Producing an item that will provide the customer with satisfaction when used, and also provide en-

joyment for the activity

Produce a product which is affordable, keeping production cost to the minimum amount possible

Design compactness into the product, bearing in mind storage issues

Ensure the designed product meets a specific need, targeting specific areas of the body to ensure

good results.

(http://exercise.about.com/cs/exercisegear/bb/home_fitness.htm—accessed 14/1/2012)

Functional fitness training is a large trend growing within the industry. This type of training uses

strength training to improve performance for activities of daily living. Exercises that mimic actual job

tasks or other activities will help improve balance, coordination, strength, and endurance. This type of

exercise compliments the idea that a healthy lifestyle is no longer just a luxury experience but some-

thing of a necessity. As the population ages, the risks associated with neglecting health and wellbeing

are motivating more individuals to invest in a health exercise regime.

There are obviously many benefits in being able to exercise, such as reduced risk of premature death,

reduced risk of depression, helping to control weight etc. There is therefore a clear need for some

home exercise equipment.

(http://www.slideshare.net/drkaufman/ihrsaeconomic20091—accessed 14/1/2012)

(http://www.slideshare.net/tinaelaine/fitness-industry-analysis-3053375—accessed 14/1/2012)

By considering the need to develop a product for a specific exercise/area of the body and the consider-

ation that functional fitness training is the most efficient way of achieving improved fitness, strength,

balance and coordination for daily life I have chosen to concentrate my product development on the ar-

ea of physiotherapy and the development of a product which can be used the home for simple exercise

to improve movement and strength.

Physiotherapy

Physiotherapy is aimed at maximising the quality of life and movement potential within spheres of pro-

motion, prevention, diagnosis, treatment/intervention, and rehabilitation. One of the most recognised

areas of physiotherapy is the work to prevent the loss of mobility before it occurs by developing fitness

oriented programmes to develop and restore maximum movement and functional ability throughout a

lifespan.

For exercise purposes the specialist area of physiotherapy which the product will aim to assist is the

area of geriatric physical therapy. This area is normally focused on the older adult and there are many

conditions which affect people as they grow older, such as;

Arthritis

Osteoporosis

Cancer

Alzheimer’s disease

Hip and joint replacements

Balance disorders

(http://en.wikipedia.org.wiki/Physical_therapy—accessed 14/1/2012)

Joint Manipulation

Joint manipulation is a common practice amongst physiotherapists and is aimed at one or more target

joints with the aim of achieving a therapeutic effect whilst also strengthening muscles in the specific ar-

ea. The main clinical effects of joint manipulation are;

Temporary relief of musculoskeletal pain

Shortened time to recover from acute sprains

Increase in passive range of motion

Physiological effects upon the central nervous system.

This is all easily achievable through the use of some very simple, motion exercise with the support of

appropriate exercise equipment. This form of mobilisation is known as instrument-assisted mobilisa-

tion. It is based upon established physiotherapy mobilising principles such as Mulligan’s positional fault

concept and facet joint glides. In the practice of physiotherapy instrument mobilisation the instrument

replicates mobilising and manipulative forces more accurately and reliably than other manual tech-

niques. This is a successful method and the design of a piece of exercise equipment for the home to

provide such exercises in a home setting could further improve the effect which it has on physiotherapy

patients.

(http://en.wikipedia.org.wiki/Joint-manipulation—accessed 14/1/2012)

Introduction to the Project

1


See Hand-In for Sketches


The Product

This product is aimed at those suffering from limited movement in the hip joint. To use the product the

foot is placed on the foot rest and forward pressure is applied. This will make the foot rest move for-

ward on the slide rail. The motor and electromagnetic combination will provide resistance, the user will

be able to determine this by using the arrow buttons provided on the control panel. The control panel

will automatically become active when pressure is applied to the foot rest. The pressure required will

be little as it is designed to activate as soon as a foot touches the foot rest. This operation will require a

pressure sensor placed below the rubber grip in the foot rest. The product will also require a timer cir-

cuit to be incorporated within the control unit. This will allow the user to select a desired time for the

duration of the activity, the timing circuit will then count to this particular time and a buzzer will signal

the end of the activity when the predetermined time is reached. The flow control diagrams for the oper-

ation of the timer circuit and pre-programmed options with the product can be seen in a section at the

end of the folio.

Target User

This product has a very specific target user. The aim is for the product to provide an exercise associat-

ed with geriatric physical therapy, in particular the movement of the hip joint and the strengthening of

the thigh muscles. This type of exercise directly affects those with hip replacements, arthritis, osteopo-

rosis and also those who are receiving rehabilitation treatment after having a limb amputation. This

product isn’t designed to replace the treatment carried out by a physiotherapist it is an aid which can be

used at home to maximise the impact of this kind of treatment. This product will allow the user to have

access to further exercise in a home environment as an additional option, maintaining movement and

progress between physiotherapy appointments and hopefully resulting in better long-term movement of

the joints, improved strength and a faster recovery.

How to use the product

The product will be used while the user is in a seated position. This is necessary as many of these

people, especially those with limb amputation, will have a weaker balance and so to avoid injury this

product must be used when seated. The user will be seated and the product placed on the floor in front

of them. Depending on which leg the user wants to strengthen will determine which foot is to be placed

on the foot rest. The product has only one foot rest to avoid any temptation the user may have to try to

stand to use the product, it may also be the case that the user only needs or wants to exercise one leg

in particular. The product can be placed at any distance away from the foot of the user, the user should

feel comfortable and find it easy to move the foot rest.

How much pressure will have to be applied to work the wheel-driven slider?

A typical pressure will be applied through the foot as it is used to drive the foot rest forward. The force

exerted through the foot whilst the person is seated is unknown but it can be estimated by using a per-

centage of the reaction force which is exerted against the ground whilst running. This is force acting in

the opposite direction and equal to that which is being exerted in a backward motion through the foot

every time it strikes the ground. This can be estimated as between 1083—1314 N. As all of the force

required for running will not be present when the person is exercising while seated I will therefore as-

sume that only half of the person’s body mass acts through the leg while seated therefore the force ap-

plied through the foot will 50% less than the force experienced while running. By taking the higher val-

ue of 1314 N and finding 50% this gives an estimated applied force of 657 N and this is value which I

will use for all further engineering design analysis. (http://hypertextbook.com/facts/1999/

Sarabirnbaum.html—accessed 14/1/2012)

Will the pressure asserted on the foot rest have an adverse frictional affect on the material? If so how

much wear will this create on the material?

The friction to be considered within this component is dry friction. This will occur because of the wheel-driven slider mechanism, where the inside surface of the drive wheel connects with the outer surface of the slide rail.. This will introduce kinetic friction into the component, occurring when the wheel rotates. This may lead to wear within the material which can cause performance degradation as well as damage to the components. As the component this will be incurring in will be made from a metal or a composite material I will need to look at the effects of wear within these materials. http://en.wikipedia.org/wiki/Friction—accessed 14/1/2012

The most probable type of wear to occur will be two body abrasive wear. This occurs when the surface roughness/peaks of one metal cut into another, this will occur during the rotating action needed in the wheels. The main causes of this are lack of lubrication of excess surface roughness, therefore a way of lubricating the handles will need to be included within the design and the surface finish of the material has to be considered to keep the wear on the material to a minimum. http://www.machinerylubrication.com/Read/1375/wear-modes-lubricated—accessed 14/1/2012

How big does the foot rest need to be and what is the range of motion required?

Some useful anthropometric dimensions required for the design of the product are;

Sitting height— Men 95th percentile; 970, Women 5th percentile; 800

Foot length— Men 95th percentile; 290, Women 5th percentile; 220

Foot Breadth— Men 95th percentile; 110, Women 5th percentile; 80

Knee height— Men 95th percentile; 580, Women 5th percentile; 460

Range of knee motion—no more than 45°

All of these measurements are in mm and are based on British persons between the age of 19 and 65.

(http://roymech.co.uk/Useful_Tables/Human/Human_sizes.html—accessed 14/1/2012)

(http://ptjournal.apta.org/content/65/7/1055—accessed 14/1/2012)

A range of knee motion of 45° combined with a knee height of 580 mm then this translates into a hori-

zontal movement of a distance of 580mm. I chose to use 45° due to experiments and motion analysis

carried out while seated. It was noticed that the leg could not extend to an angle of more than 45°

while also simultaneously keeping the foot on the ground.

Will sections of material with a high frictional property be required to provide the user with grip?

Many materials will typically have a coefficient of friction value between 0.3 and 0.6, however rubber in contact with other surfaces can yield a coefficient of friction between 1 and 2. This coefficient can only help in providing ease of use when considering the usability of the product. This will be the main con-tact surface between the foot and the foot rest. I am intent on not have straps as this causes hassle and so a high frictional value is required to provide the grip needed for use. This will also provide com-fort as a soft material has been provided on which the foot will rest.

Embodiment Design 1

5

http://en.wikipedia.org/wiki/Friction

http://en.wikipedia.org/wiki/Friction

http://www.machinerylubrication.com/Read/1375/wear-modes-lubricated

http://www.machinerylubrication.com/Read/1375/wear-modes-lubricated


See Hand-In for Sketches


As was explained during the embodiment design section, I am going to assume that only 50% of the

1314N force exerted through the foot while running is exerted through the foot whilst sitting using this

product. Bearing this in mind I have devised a free body diagram for this product as shown opposite

.

The diagram shown above is the free body diagram modelled as a beam on Dr. Beam software to cal-

culate reaction forces and shear force and bending moments.

F1 = The resultant force from the foot

(Although the maximum extension of the knee, in order to use this product, is 45° I am going to assume

that the force exerted from the foot all acts vertically downward as it makes it easier to model the sys-

tem on Dr Beam to calculate values for the other forces. This will not underestimate the force being ap-

plied to the product as in real life the force will be less than what has been modelled here so there will

be a factor of safety within the design.)

F2 = The reaction force acting in the opposite direction from the horizontal component of the foot force

to prevent the product from sliding along the floor.

RA and RB = The reaction forces where the metal framework interacts with the ground, The oppose the

forces acting vertically downward on the frame.

W = The weight of the sliding mechanism components acting on the metal framework.

(This provides a considerable force and therefore cannot be ignored.)

F1 is shown as a distributed load on the beam model within the Dr. Beam software as the force from the

foot is applied along the whole base of the foot plate component. The force is only shown to be acting

along a distance of 60mm and not the length of 290mm of the foot plate as the foot plate is attached to

the slide rail via a square section which is only 80mm long. The force along this 80mm component will

have the greatest bending moment and shear stress when the sliding mechanism is in the middle of the

slide rail, so this is how I have modelled the product for the force analysis to ensure the section sizes

chosen are correct for the maximum bending moment.

Above is the free body diagram for the whole framework assembly of the forward and backward step-

per.

The above diagram shows the outcome of a bending moment diagram calculated by Dr. Beam.

Finding W

Weight of foot plate = 0.35751066 kg x 9.81 = 3.507179 N

Weight of square box section = 0.0388653 kg x 9.81 = 0.38126 N

Weight of slide mechanism plate = 0.130094 kg x 9.81 = 1.276222 N

Weight of wheels = (0.013941157 kg x4) x 9.81 = 0.54705 N

Weight of slide rail = 0.39884752 kg x 9.81 = 3.91269 N

Weight of box section = 0.4897 kg x 9.81 = 4.80395 N

Box section support = 0.32858994 kg x 9.81 = 3.22346 N

Therefore W = 3.507179 + 0.38126 + 1.276222 + 0.54705 + 3.91269 + 4.80395 + 3.22346 =

17.61 N

Maximum Bending Moment = 0.0081 kNm

Maximum Shear Force = 0.0288 kN

Force Analysis 1

7

Max. shear force

Max. bending moment

80


My next aim through force analysis is to identify the minimum standard box section dimensions which

should be used for the foot plate support. This component is attached to the slide mechanism plate

which houses the wheels and on top has the foot plate where the user places their foot. Below is the

free body diagram for this component. (The free body diagram is based on the use of a 20mm x 20mm

square box section. This will not affect the force values given by Dr. Beam and the minimum standard

box section size will be calculated from these.)

The diagram below is the force calculation results

from Dr. Beam.

Maximum bending moment = 0.1Nm Maximum shear force = 14.9N

Using the maximum bending force I can now use the bending equation to determine the minimum sec-

tion dimensions suitable for the product.

σb = M/Z

Z = I/Y

I am going to use aluminium for this chosen component and to solve the equation for the bending

stress I first need to identify the maximum allowable bending stress.

Aluminium 6061 –T4—yield stress = (110 MPa) [1]

Factor of safety = 2

Maximum allowable stress (σa) = 110 x 106 / 2 = 55 x 10

6

σb = σa

Z = M/σa

Z = 0.1 / 55 x 106

Z = 2 x 10-9

m3

The standard section chosen must therefore have a Z value of 2 x 10-3

cm3 or above. The table below

highlights my chosen section sizes.

Z = I/Y = 0.88/1

= 0.88 cm3

The table above shows the chosen standard, square box section I have chosen for this component. It

is clear to see from the calculation that this size of section has a Z value of 2 x 10-3

cm3 or above. I

haven’t chosen a section which has exactly this Z value as choosing a higher value gives a higher fac-

tor of safety meaning the design is less likely to fail. My chosen section is therefore a 20mm x 20mm

square box section with 2.6mm wall thickness. The material used is Aluminium 6061-T4 as this has a

suitably high yield stress value. The component will therefore not fail but it is also light and cheap for

production.

The table has been taken from a data booklet issued in Design 2, May 2011

[1] http://en.wikipedia.org/wiki/6061_aluminium_alloy—accessed 2/12/11

Force Analysis 2

8

Max. shear force

Max. Bending Moment


The diagram below is the free body diagram for the wheel-driven slider mechanism for the forward and

backward stepper. This force analysis was primarily to determine the forces acting on the wheel which

would be driving the mechanism.

As you can see from the diagram there is a distributed load of 657N acting across the central 20mm of

the plate which houses the wheels. This comes from the force of the foot being transmitted through the

20mm x 20mm square box section which was detailed on the previous page. W represents the weight

of the two components, the square box section and the foot plate, which are fixed on top of the plate.

This results in a reaction force at the centre of the wheels, where they will be fixed, of 8.5N. Below is

the Dr. Beam graph showing bending moments and shear stress.

Maximum bending moment = 0.2 Nm Maximum shear force = 8.5 N

The free body diagram below shows the front view of the slide rail component. The main aim of this

analysis was to determine the forces acting on this component.

This diagram shows a distributed load acting

across the top surface of the component.

In reality this would probably not be the case but there will be a force acting on the top surface. To en-

sure the force analysis of later components is accurate and values are not underestimated I have de-

cided to place this distributed load in the diagram and use it throughout the force analysis. By doing

this the values obtained will be overestimated and will add to the factor of safety for failure built into the

design of the component. Below is the outcome from the Dr. Beam analysis software.

In this component W again represents the weight of the components placed on top of this compo-

nent. This adds force to the component which must be considered. Compared with the other

components which have been analysed, the slide rail has a relatively high shear force but negligi-

ble bending moment. These conditions only apply to the cross-section of this component. The

forces and bending moments occurring along the length of the slide rail will be determined on the

next page. The results gathered from this analysis will be used later in a finite element analysis

using the Pro Mechanica application available on ProE.

Force Analysis 3

9

Max. Bending moment

Max. Shear stress

Max. shear force

Note; negligible bending moment


A standard size for the section on which the slide rail is attached needs to be established. This will be

done in the same way as shown on page 10. The Dr Beam software has modelled the section as a

beam with the force exerted by the foot and the weight of the other components placed on the section

illustrated as shown below.

In order to find the sec-

tion size needed, the Engineer’s bending formula is used.

Maximum bending moment— 0.2Nm Maximum shear force— 19.7N

σb = M/Z

Z = I/Y

As with the component shown on page 10 I am going to use a grade of aluminium for this component,

so the allowable stress for this component is;




6

σb = σa

Z = M/σa

Z = 0.2 / 55 x 106

Z = 4 x 10-9

m3

The standard section size chosen for this component must therefore have a Z value of 4 x 10-3

cm3 or

above to prevent the section failing in bending. The table opposite details some of the standard square

section sizes available for use and highlights my chosen section size for this particular component for

the forward and backward stepper.

Z = I/Y = 4/1.5 = 2.667 cm3

Y is the perpendicular distance to the neutral axis and is equal to 1.5 cm in this instance. It is clear to

see from the calculation that this size of section has a Z value of 4 x 10-3

cm3 or above. I haven’t cho-

sen a section which has exactly this Z value as choosing a higher value gives a higher factor of safety

meaning the design is less likely to fail. My chosen section is therefore a 30mm x 30mm square box

section with 3.2mm wall thickness. The material used is Aluminium 6061-T4 as this has a suitably high

yield stress value. The component will therefore not fail but it is also light and cheap for production.

The table has been taken from a data booklet issued in Design 2, May 2011

[1] http://en.wikipedia.org/wiki/6061_aluminium_alloy—accessed 2/12/11

Design For Bending

To ensure the component doesn’t fail in bending, I have checked the bending stress generated using

the Engineer’s bending formula;

σ= MY/I

= (0.2 x 0.015) / I

I = bh3/12

= (0.03 x (0.03)3)/12 = 6.8 x 10

-8

(0.2 x 0.015) / 6.8 x 10-8

= 4.41 x 104

Shear stress for material is 110 x 106

Factor of safety for bending = 110 x 106 / 4.41 x 10

4


This shows the component can take almost 2000 times more stress than what it is taking in this product

before the shear stress of the material is reached and the component fails through bending.

Force Analysis 4

10

Max. Shear Force

Max. Bending Moment


The final force analysis required is the analysis of the section size and forces imposed upon the metal

support frame which holds all of the other components in place and provides the structural support re-

quired for the use of the product. This has been done using free body diagrams and the use of the Dr

Beam software combined with the engineer’s bending formula. The diagram below is the free body dia-

gram for this component.

Although it is unclear from the diagram above, the forces acting on this component act in two perpen-

dicular directions, i.e. horizontally and vertically. To find the maximum bending force acting on this

component two bending force diagrams will be required from Dr Beam, they will be split into the hori-

zontal and vertical force directions. The diagram below shows the results for the vertical forces.

Maximum vertical shear force = 25.2N

Maximum vertical bending moment = 1.9N

The diagram below shows the resulting graphs for the horizontal forces acting on this component.

Maximum shear force = 543N Maximum bending moment = 23.1N

To find the maximum bending force acting on the component, both bending moment diagram have to

be combined. This is done by taking the bending moment values at the exact distance where each

force acts from both diagrams and combining taking the square root of both of the values squared.

This is done below.

The total maximum bending moment;

√ (23.12 + 0.5

2) = 23.105N—This occurs at the critical section which is 0.042m from the left

To find the section size needed;




6

σb = σa

Z = M/σa

Z = 23.105 / 55 x 106

Z = 4.2 x 10-7

m3

The standard section size chosen for this component must therefore have a Z value of 4.2 x 10-7

cm3 or above to prevent the section failing in bending. Using a data table from www.cim.mcgill.ca/

~paul/HollowStruct.pdf—accessed 14/1/2012, the Z value for a circular hollow section of 27mm is

0.0089 m3. I can therefore make the section size smaller than this, if I make the section size

10mm, that is roughly 3 times smaller, assuming the Z value is therefore also 3 times smaller this

gives a Z value of 0.00296 m3. This therefore means that the section will not fail in bending. I feel I

can make this assumption as the Z value required is so small that assuming the Z value for a

10mm section size and comparing it with another section size with an available Z value is appropri-

ate. In order to ensure failure does not occur I have chosen a large wall thickness for this section.

Force Analysis 5

11

Max. Shear Force

Max. Bending Moment

Max. shear force

Max. bending moment


The section sizes for some of the key components in the design have been chosen with bending in

mind but these sections can also fail through buckling, so to ensure the section sizes chosen are also

suitable when buckling is considered I have carried out the following calculations.

The material being used in these sections is Aluminium 6061 –T4 with the following values;

Young’s Modulus = 66GPa and Tensile Yield Strength = 110MPa [2]

[2] - http://en.wikipedia.org/wiki/6061_aluminium_alloy—accessed 27/12/2011

When considering buckling as an entity it is also necessary to identify a Leff value. As both of the sec-

tions being considered are fixed at both ends they will therefore have the same Leff value which is given

in the table shown below;

In the table above, taken from an Engineering Design lecture presentation accessed via myplace,

shows the different possible Leff values. The one which represents the column having both ends fixed

is ( c ) therefore giving me an Leff value of 0.65L as the minimum recommended value, where L is the

length of the section.

The slenderness ratio for each section will need to be checked against the Johnston criterion for buck-

ling and as the same material is used for all section sizes considered, this value will be the same for

each. The Johnston criterion is as follows;

Ω√(2E/Sy) E = Young’s modulus, Sy= Tensile yield stress

= Ω√((2 x 66x109)/110x10

6) = 108.83

Use of Dr Frame 3D software

Buckling equations require the area and second moment of inertia values of the sections to be known.

As this is complicated to calculate by hand I modelled each of the section sizes using the Dr Frame 3D

software and simply read-off the values which were automatically generated for me. This was much

easier, saved time and was more accurate than calculations carried out by hand could have been.

Buckling

Mechanical failures in products do not always come from loads which exceed the yield strength of ma-

terials. It can be due to a situation in which the structure, once disturbed in some way, cannot restore

itself to its original state. This is instability within the material and the structure and can be assessed

through different forms of buckling calculations. Euler and Johnston buckling are the most common

forms and this is what has been used in the design of the forward and backward stepper. When col-

umns buckle they do not bend slightly at first and then collapse as the load increases, rather, the fail-

ures are cataclysmic and the column can suddenly collapse without warning when a critical load is

reached. This is why buckling must be considered as an addition to considering bending when design-

ing. (http://classes.myplace.strath.ac.uk/file.php/15925/2011_Lecture_5_-_Euler_bucklingIntro.pdf)

Buckling in the Rectangular Tube sections

For the section size— 30mm x 30mm x 3.2mm (thickness) L = 630mm

A = 3.17x10-4

m2 I = 3.62x10

-8m

4

Leff = 0.65L Therefore Leff = 0.65 x 0.630 = 0.4095 m2

ρ = √(I/A) = √((3.62x10-8

)/(3.17x10-4

)) = 0.01686m2

Slenderness Ratio = Leff/ρ = 0.4095/0.01686 = 38.32033102

The slenderness ratio value is less than the Johnston criterion value which was calculated opposite

therefore this column acts as a Johnston column so the buckling equation used is as follows;

PCR/A = Sy - (SY2/(4Ω

2E))(Leff/ρ)

2

PCR/A = (110x106) - ((110x10

6)2/(4Ω

2(66x10

9))(38.32033102)

2

PCR/A = 103180693.6

PCR = 32708.278988 Pa

Factor of Safety = PCR/P = Critical Load/Load on member = 32708.278988/658.8 = 49.648

(the load on the member has been taken from the total forces acting on the member in the free body

diagram on page 11)

This member has a factor of safety of 49 and will therefore not fail in buckling. The factor of safety for

the 20mm x 20mm rectangular tube has been found in exactly the same way as shown above and the

critical load for this section is;17996.53346Pa and the factor of safety is 27.4. A factor of safety of 27

shows this member will also not fail due to buckling.

Buckling in the Circular Tube Section

For the section size— 30mm x 3.6mm (thickness) L = 630mm

A = 6.04 x 10-4

m2 I = 4.65 x 10

-10m

4

Leff = 0.65L Therefore Leff = 0.65 x 0.245 = 0.1593 m2

ρ = √(I/A) = √((4.65x10-10

)/(6.04x10-4

)) = 8.774x10-4

m2

Slenderness Ratio = Leff/ρ = 0.1593/0.000877 = 181.55

The slenderness ratio value is more than the Johnston criterion value which was calculated opposite

therefore this column acts as an Euler column so the buckling equation used is as follows;

PCR = Ω2EI/Leff

2

PCR= (Ω2 x (66 x 10

9) x (4.65 x 10

-10))/0.1593

2

PCR = 11936.172 Pa

Factor of Safety = PCR/P = Critical Load/Load on member = 11936.172/657 = 18.17

(the load on the member has been taken from the total forces acting on the member in the free body

diagram the previous page.)

This member has a factor of safety of 18 and will therefore not fail in buckling.

Design for Buckling

12


Welding

There are two welded joints in the structure of the frame for this product. This occurs where the sup-

port plate for the 30mm x 30mm rectangular tube section comes into contact with the circular tube sec-

tion which creates the support structure for the entire product. The only way of joining these two ele-

ments is through welding. To calculate the thickness of the weld required it is assumed that a fillet weld

will be used. An E43 electrode of S355 grade steel is being used. The table below has been used to

find the required shear stress yield value for the electrode material to be used in the calculations;

The correct shear stress yield value has been identified

from the table, however, the required approximate value

for a fillet weld is 0.75 x 250 = 187.5 N/mm2

Weld length = 36mm


The maximum shear force placed on the component, giv-

en in the graph on page 12, is 19.7N. When considering the factor of safety the force placed on the

component is 19.7 x 2 = 39.4N

σs = F/tL 187.5 = 39.4/t x 36

t = 0.005837037mm

S = t/cos45° = 0.005837037/0.707

S = 0.008256064mm

This calculation suggest a weld thickness of 1mm, however welds need a minimum size of s = 10mm

therefore the suggested thickness of the weld for this design is 10mm. (3) (http://

classes.myplace.strath.ac.uk/file.php/15925/Supplemental_Information_for_DM303_course_work.pdf)

Fastening

For all fastening equations I have assumed that a steel 12.9 bolt will be used with a factor of safety of 2

being implemented.

The table

above shows that the tensile yield strength of the bolt being used is 1200MPa. (3)

The fastening equation being used is;

D2/4 = τ / N x Ω x σY

D is the diameter of the bolt required, τ is the shear stress acting on the member multiplied by

the factor of safety, N is the number of bolts required and σY is the tensile yield stress taken

from the table above.

Foot Plate Support

Shear Force = 14.9N Factor of Safety = Shear force x 2 = 29.8N = τ

Number of Bolts = 2 σY = 1200MPa (N/mm2)

D2/4 = 29.8 / 2 x Ω x 1200

D2 = 0.01581 D = 0.126mm

The suggested diameter for a bolt in the foot plate support is 1mm. I am going to use a 3mm diameter

in the design as this adds an extra factor of safety and provides a better overall look to the design.

Sliding Mechanism Plate

Shear Force = 0.5N Factor of Safety (τ) = Shear force x 2 = 1N


D2/4 = 1 / 4 x Ω x 1200

D2 = 0.0002653 D = 0.0163mm

This calculation also suggests using a 1mm diameter bolt, however, for the same reasons a 3mm bolt

will be used.

Slide Rail

Shear Force = 10.7N Factor of Safety (τ) = Shear force x 2 = 21.4N


D2/4 = 21.4 / 9 x Ω x 1200

D2 = 0.002523 D = 0.0502mm

The same result has been calculated for this component, again I am going to use a 3mm diameter bolt.

Load Supporting Box Section



D2/4 = 39.4 / 9 x Ω x 1200

D2 = 0.004645 D = 0.0682mm

Again for this component I will use a 3mm diameter bolt.

Rectangular Tube Support Frame



D2/4 = 39.4 / 9 x Ω x 1200

D2 = 0.004645 D = 0.0682mm

Fastening and Welding

13


The axial supporting the gear for the chain and the disk with the magnetic bars will be rotating every

time this exercise equipment is used, therefore to aid movement and limit wear two ball roller bearings

are required. One floating and one locating ball roller bearing will be required for the best outcome. I

have chosen to use a ball roller bearing as there is a small axial load, it is cheap to keep the overall

cost of the product to a minimum and I don’t feel that noise caused by the equipment is a major issue to

the user. For the following bearing selection calculations it is assumed that Fa (the axial load) is equal

to approximately 30N, coming from the weight of the gear and disk placed on the axil, and Fr (the radial

load) is equal to the forward component force from the foot which drives the sliding mechanism, this is

equal to 675N.

Table Q2.2 – Typical Bearing Design Life (Hours) (Tables were taken from an engineering design

lecture via myplace http://classes.myplace.strath.ac.uk/course/view.php?id=15925)

The forward and backward stepper can be classed as both an intermittent-domestic machine and a ma-

chine which is used for short periods. The bearing design life hours therefore ranges between 300 and

8000 hours. As 3000 hours is common to both classes of machine and fails in the middle of the range

it is therefore suitable to choose 3000 as the life hours value to be used in the bearing selection calcu-

lations.

Static Load Bearing Capacity

The static load rating of a bearing is;

C0 = S0P0 Where S0 is the static safety factor and P0 is the equivalent static bearing load

S0 can be found from the table below;

S0 for this design

will be 0.5 and P0

can be found by the ration of Fa to Fr

Fa/Fr = 30/657 = 0.0457 Therefore Fa/Fr ≤ 0.8 so P0 = Fr which is equal to 657N

Therefore C0 = 0.5 x 657 = 328.5N

Dynamic Loading

The equivalent dynamic load P also needs to be calculated as the bearing is subject to both static and

dynamic loading.

P = XFr + YFa where X is the radial factor and Y is the axial factor. To find values for X and Y the ratio

of Fa/Fr needs to be used again, as before this equals 0.0457, this is ≤ 0.35 and using this information

values for X and Y can be taken from the table below;

As Fa/Fr is ≤ e X = 1 and Y = 0. For the remaining dynamic

loading calculations it is to be assumed that the rpm value of

the axle is 500 and k = 3 for a ball roller bearings. This will help

calculate the dynamic load rating c.

C = PL101/k

L10 = rpm x 60 x life (hours)/1,000,000

L10 = 500 x 60 x 3000/1,000,000 = 90

P = 1x 657 x 0 = 657N

C = PL101/k

= 657 x 601/3

= 2572N

Bearing Selection

Single Row Deep Groove Ball Bearing selection data sheet.

The chosen ball roller bearing has a designation of E2.627-2Z/C3 as given by the supplier http://

www.skf.com/portal/skf/home/products?newlink=1_1_16&lang=en. The component has a mass of

0.013kg and the dimensions are as follows;

Bore diameter = 7mm, Outer diameter = 22mm and the Bearing width =7mm. The dynamic load rating

for this component is 3.32kN and the static load rating is 1.37kN. This is more than what was calculat-

ed above to ensure the component does not fail during use. To check that the L10 value is still accepta-

ble with this particular bearing the given dynamic load rating has been placed in the equation;

L10 = (C/P)k = (3320/657)

3 = 129 million revolutions

This is the minimum bearing required, the one chosen will have a bore diameter of 10mm to accommo-

date the axle on which the bearings are attached.

Bearing Selection

14

Machine Usage Hours

Intermittent- domestic machines 300-3000

Short periods- hand tools, construction machines 3000-8000

High reliability for short periods- lifts, cranes 8000-12000

8 h/day partial use gears, motors 10000-25000

8 h/day full use machine tools, fans 20000-30000

Continuous use 40000-50000

Noise Unimportant Normal Running Quiet Running

Ball Roller Ball Roller Ball Roller

Smooth loading 0.5 1 1 1.5 2 3

Normal loading 0.5 1 1 1.5 2 3.5

Shock loading >=1.5 >=2.5 >=1.5 >=3 >=2 >=4

Fa/Fr >0.8 Po = 0.6Fr + 0.5Fa

Fa/Fr 0.8 Po = Fr

P= XFr + YFa

Co/Fa e Fa/Fr e Fa/Fr > e

X Y X Y

5 0.35 1 0 0.56 1.26


ProE Mechanica Analysis

15

The diagram to the left details a design study analy-

sis carried out on proEngineer on the foot rest com-

ponent for the product. The results show that the

maximum stress experienced by the component is

5.996 x 102 N and this occurs around one of the

holes in the surface. The maximum displacement

of the component due to the force exerted is 9.736

x 10-1

mm. Regarding the material properties this

mechanical analysis has shown that the stress ap-

plied will not cause the component to fail. The re-

sults show that the maximum displacement occurs

around the front curve of the component, this is due

to the force being applied to the front of the compo-

nent, driving the wheel-driven slider mechanism for-

ward.

The diagram on the right details the design study analysis for

the slide rail element of the product. This is a critical compo-

nent as it enables the movement of the foot rest. The maxi-

mum displacement experienced is 6.189 x 10-2

mm, and the

maximum stress occurring is 5.197 x 101 N. From the re-

sults picture it is clear to see that this top value of stress al-

most doesn’t occur anywhere on the component however

when considering whether the component is suitable for use

or not then this higher value of stress must be used. The

results show that this component is also suitable for use.

The other critical components were also analysed using the

same method and were found to be suitable for use.


Cutting

The standard section sizes which create the overall form and shape of the exercise equipment will need

to be cut to the correct length for the product. A rubber tread will also need to be cut to be placed into

the foot plate to provide grip for the base of the foot while the user is moving the plate back and for-

ward.

Laser cutting— A laser beam, focused by a lens to a focus spot, in the 100μm to 330μm will allow for

thicker aluminium section cutting. Coaxial gas surrounds the converging laser beam to keep the optics

clean and improve the cutting process. Gases such as oxygen and air are used to promote cutting of

ferrous alloys and cellulose materials. Air is often used with plastics and fabrics. High-pressure inert

gas is used when cutting some metals to leave an un-oxidized edge with little dross. A pulsed nd:YAG

laser is the suggested laser type for this manufacturing process and cuts thicker than 25mm can be

achieved. The same machine can also be used for laser drilling and laser welding.[3]

Blanking—Blanking is the operation of punching, cutting or shearing a predetermined shape from a

sheet material by die cutting the outer shape of the part.[4] A suitable blanking production process for

this project would be a simple ‘cookie cutter’ type die which simple cuts out the shape of the rubber grip

for the foot plate. This would save on the cost of buying expensive equipment to run high speed dies

which turn at over 1000 revolutions per minute. This would simply be an unwise investment as low

numbers of this product will be required due to the small size of the niche market which the product is

being aimed at. This will result in lowering the overall retail price of the product as the manufacturing

cost will be kept to a minimum by keeping the manufacturing process to simple necessary operations.[5]

Design Considerations

Laser cutting—Burrs are quite small in relation to blanking and shearing. They can be almost eliminat-

ed when 3D lasers are used and further eliminate the need for secondary deburring operations. As la-

ser operates by melting the material in the beam path. Materials which are heat treated will become

case hardened at the cut edges. This may be desirable if the hardened edges are functionally desirable

within the finished product, however if secondary machining operations are required, such as threading,

then hardening is a problem.[6]

Blanking—Corners should have a radius of 0.5 x the material thickness or 0.4mm, whichever is greater.

Sharper corners can be produced however this also increase the die maintenance costs and produces

more burrs during the manufacturing process. Slot or tab widths should be 1.5 x the thickness of the

material. On cut offs it is best to avoid full radii across the width of the material as it results in a feather

burr running the length of the radius. A square cut off produces the best result. If a radius is necessary

then and angle-blended radius is the best option. [5]

[3] - http://www.gsiglasers.com/LaserProcesses.aspx?page=18—accessed 30/12/2011

[4] - http://www.advantagefabricatedmetals.com/blanking.html—accessed 30/12/2011

[5] - http://www.efunda.com/processes/metal_processing/stamping_blanking.cfm—accessed

30/12/2011

[6] - http://www.efunda.com/processes/metal_processing/laser_cutting.cfm—accessed 30/12/2011

Aluminium Die Casting—The die casting process is similar to the permanent mould casting process ex-

cept in die casting the metal is injected into the mould at high pressure. This results in a more uniform

part with good surface finish and good dimensional accuracy. This process allows post-machining to be

eliminated for most parts speeding up the manufacturing process and reducing production costs. Alu-

minium, alloyed with silicon and copper, is cast at 650°C. Silicon increases the melt fluidity and reduc-

es machinability whereas copper increases hardness and reduces ductility. Minimum wall thicknesses

and draft angles for aluminium die casting are 0.9mm and 0.5° respectively.

Design Considerations—It is best to design cast components with uniform wall thicknesses and cores of

simple shapes. Heavy sections cause problems in cooling and trapped gases result in porosity. All cor-

ners should have sufficient radii to avoid stress concentration. Draft allowance should be applied to all

for the release of parts, this is generally between 0.25° and 0.75° per side depending on wall thickness.

[11]

Tube Bending—Tube bending is used to permanently form pipes and tubing. The tube is loaded into a

pipe bender and clamped between two dies. Mechanical force is used to push the tubing against a die,

the tube then conforms to the shape of the die. For the shape required for this manufacturing process,

A mandrel will be placed inside the tube to prevent collapsing. The tube will also be held loosely in

place by a wiper die to prevent any creasing during stress. The wiper die is normally made from a soft-

er alloy to avoid scratching or damage to the material being bent.[10]

Bending—Bottoming—Bending is done using press brakes. Press brakes normally have a capacity be-

tween 20 and 200 tons. Larger and smaller presses are available for specialised operations. Program-

mable back gages and multiple die sets which are currently available can make a very economical pro-

cess. Bottoming is the bending process where the punch and the work piece bottom on the die. This

allows for a controlled angle with very little spring back. The inner radius of the work piece should be a

minimum of the equivalent material thickness in the case of bottoming. This is therefore the most suita-

ble bending process for the sheet aluminium which supports the 30mm x 30mm square box section.[7]

Welding

Welding is a process which permanently joins two or more metal parts. Aluminium can be welded using

high power, oxide free surfaces. However, cleaning the surfaces to be oxide free adds to the cost of

manufacture. The thickness of parts to be welded should be equal.[8] Tungsten Inert Gas Welding

(TIG) should be used during the manufacturing process. An arc is struck between the tungsten elec-

trode and the sheet metal to be welded. An inert gas shields the arc from the ambient in order to pre-

vent oxidation. This is a cheap, quick form of welding which produces less spatter than that which oc-

curs during MIG welding which is the reason for choosing this particular method of welding.[9]

[7] - http://www.efunda.com/processes/metal_processing/bending.cfm—accessed 30/12/2011

[8] - http://www.efunda.com/processes/metal_processing/welding.cfm—accessed 30/12/2011

[9] - http://www.efunda.com/processes/metal_processing/welding_inertgas.cfm—accessed 30/12/2011

[10] - http://en.wikipedia.org/wiki/Tube_bending—accessed 31/12/2011

[11] - http://www.efunda.com/processes/metal_processing/die_casting.cfm—accessed 31/12/2011

Manufacturing 1

16


Injection Moulding

Cost Considerations—Many factors add to the cost of an injection moulded part, including the material

selection, the section complexity and size of the part, the complexity of the design, the finish and ap-

pearance required etc. When considering cost in the design of a part for injection moulding then there

are two key areas to consider, the design of the part to minimise production costs and the design of the

part to minimise tooling costs. To minimise production costs the aim is to keep the section of a part as

thin as possible which results in shorter moulding cycle times and also reduces the material content in

the design. By keeping mould tools simple and if possible combining several parts into one moulding

run then the cost of tooling will also be minimised.

Design Development—The end user requirements for the plastic components in this products are pri-

marily for aesthetic appearance. They will not be subjected to heavy loading and the only functional

role they will play is providing a place in which to put a piece of rubber material in order to prevent the

product from sliding along a surface during use. However, legal requirements such as flammability and

dimensional standards still need to be incorporated and it is essential that recyclability is considered as

disposal and re-use is key at the end of the product’s life expectancy.

Design for Precision—On cooling design mouldings can shrink which can cause problems later in as-

sembly. A thick section will either sink forming a curved surface or voids will be created in the material

mass. To try and prevent this, wall thicknesses for unreinforced plastics are typically kept between

0.5mm and 5mm. Another defect which can occur during cooling is warping. This occurs when the

part shrinks unevenly and cause temporary and permanent dimensional changes. Wall thickness is al-

so a critical factor in precision manufacturing. Thinner sections have a shorter cooling time and so al-

lows for less correction when in the mould or by jigging. Varying thickness within one moulded part al-

so leads to distortion due to different shrinkage occurring during cooling. To accommodate for this a

gradual transition between different sections should be used.

Design for Appearance—Various surface defects can occur during injection moulding. Sink marks oc-

cur over projections such as ribs and bosses. This is due to the localised thickening of the section

which results in above average shrinkage. This defect can be minimised by keeping ribs and bosses to

50% of the wall thickness of the material. The defects can also be disguised by the use of a textured

surface finish and styling. Weld lines occur when two flow fronts meet during the process. This defect

can be minimised through good tool design however problems occur where there are holes in the part,

the material must then flow around the hole and weld on the opposite side. This results in small weld

lines adjacent to the hole and possibly also flow lines where the smooth passage of the material in the

mould has been interrupted. This can cause physical weakening and can appear as a crack on the

surface of the finished part. Burning occurs where there is a lack of sufficient venting at the edge of a

mould. As a direct result burn spots will appear due to the effect of diesel. Air traps may also occur

around ribs and sharp transitions this then requires venting around ejector pins and in other ways in the

mould design. Voids are air bubbles which have formed in the material due to section thickness. Alt-

hough it is only visible in clear materials is can cause weakening of the component and is caused by

excessive shrinkage. Excessively thick section sizes should therefore be avoided for the best outcome,

materials which are less prone to voids should also be used.

Design for Assembly—Press and Snap Fits—When designing for assembly a mismatch in alignment of

the parts being joined should be avoided. Press fits rely on an interface between the two parts to keep

them joined. The degree of interface between the two parts is critical, too little and the joint will be

loose, too great and the joint becomes difficult and the material will be overstressed. This joint is there-

fore only viable when the manufacturing tolerances are attainable and can be maintained. Also where

different materials are involved then changes in temperature may also affect the fit. Snap fits are ideal

for an injection moulded part as they are economical through their nature of being moulded and they

are easy to recycle as no metal inserts or adhesives are involved in manufacturing them. This joining

method is not suited to parts which require repetitive assembly and they can be damaged easily by mis-

handling especially if the material being used is brittle.

Design for Mouldability—Ribs—Ribs are used to increase stiffness or strength within a component with-

out increasing the overall wall thickness. Ribs however can increase warping and enduce appearance

problems. To reduce the risk of sink marks appearing on the surface then the rib thickness should be

no more than 50% of the adjoining wall thickness. To reduce the risk of stress, filling and ejection prob-

lems the height of the rib should not exceed more than three times the adjoining wall thickness. If more

strength is required then more ribs should be added rather than increasing the height of a rib. A mini-

mum radius of 25% of the adjoining wall thickness should be incorporated at the bottom of the rib. Ribs

are most effective when they are placed down the length of the area which is subject to loading. The

spacing between each rib should be twice the nominal wall thickness. A draft angle of at least 0.5°

should be included to aid with the release from the mould. Support ribs are used to reinforce corners,

sides and walls and different design considerations should be included. The thickness of a support rib

should be between 50% and 70% of the thickness of the wall of the component. The minimum distance

between faces of ribs should be at least twice the wall thickness of the component. The minimum

length of the face of a support rib should be twice the wall thickness of the component. Significant radii

need to be incorporated at rib ends and a minimum 0.5° draft angle included. The minimum length of a

rib face which is attached to a boss should be at least four times the component wall thickness.

Design for Mouldability—Bosses and Undercuts—Bosses help to facilitate mechanical assembly and

some design considerations are; The wall thickness of a boss should be between 50% and 70% of the

nominal wall thickness; often this is insufficient and cannot withstand the stresses from an insert how-

ever a thicker wall can result in sink marks appearing on the surface. A minimum radius of 25% of the

boss wall thickness is needed at the bottom of the boss. More strength may be added through the use

of support ribs and strength can be greater increased by attaching the boss to a nearby wall using a rib.

If possible undercuts should be avoided. Mould tool should ideally have a single open and close mo-

tion however in certain circumstances mouldings with a slight undercut can be ‘bumped’ off the core.

The same affect of an undercut can be achieved by using certain design techniques, it may be neces-

sary to have side movements in the mould tool but this will increase tooling costs.

Design for Recyclability—Environmental legislation requires for all products to focus on recyclability af-

ter use. The design requirements for this need to be considered. Metal inserts should be avoided as

separation is difficult therefore making recycling uneconomic. Snap fits should be used where possible

to avoid self-tapping screws. Where possible components should be made from the same material and

grade, if different material is used then they should carry appropriate markings so the material can be

identified. Protective paints, lacquers and protective coatings should be avoided. The amount of mate-

rial used within the component should be kept to a minimum for economical reasons.[12]

[12] - http://www.rutlandplastics.co.uk/design_guidelines.shtml—accessed 1/1/2012

—accessed 1/1/2012

Manufacturing 2

17


Movement Mechanism Analysis

18

The wheel driven slider mechanism is a single axis

mechanism which is driven by four ‘v’ shaped wheels.

The ‘v’ shape on the wheel corresponds to the v shape

edge on the track, along which the wheels rotate. The

v shape provides a large surface area along which the

weight of the sliding mechanism and other attached

components can be distributed, giving a larger contact

area which results in making the mechanism more sta-

ble and less susceptible to warping or disengaging from

the rail. As the forward force component from the foot

is applied the mechanism begins to move forward. The

mechanism achieves this forward movement via rota-

tion from the wheels. The wheels are not interlinked in

any way and are therefore all free to rotate inde-

pendently from one another. The wheels located on

the left hand side of the mechanism rotate in a clock-

wise direction whereas the wheels located on the right

hand side rotate in an anticlockwise direction. This re-

sults in an overall forward movement. The sliding

mechanism is attached to a chain drive to ensure the

mechanism is not free-moving and can provide re-

sistance to use. It also ensures the foot plate returns to

At the beginning of one movement the slid-

ing mechanism sits stationary at the right

edge of the slide rail. The wheels are not

rotating at this point and the foot plate is sta-

tionary.

In the middle of the cycle the foot plate has

moved to the central point of the slide rail. A

forward force has been applied so the

wheels are rotating. The forces applied to

the components is the greatest at this point.

At the end of the cycle the foot plate will have

reached the left edge of the side rail. The for-

ward force will stop being applied and the foot

rest will return to the starting position due to

the opposite reaction force from the chain

drive.

At the beginning of the cycle the chain drive

will be stationary. The sprocket will be sta-

tionary and the elastic cord keeping tension

on the chain will be in its un-stretched state.

In the middle of the cycle the chain drive will

be moving, the sprocket will be rotating in a

clockwise direction in order to move the foot

plate. The elastic cord keeping tension on

the chain will be in its half stretched state

At the end of the cycle the chain will stop moving

once the foot plate has reached the end. The ten-

sioning cord will therefore return to its un-

stretched state returning the foot plate and the

chain to its original position.


Providing Resistance

The resistance in this product will be provided by an electro-magnetic flywheel. The flywheel will be

mounted on the same rotating shaft as the sprocket which houses the type B chain. On the flywheel

there are three bar magnets situated around the circumference of the wheel. As the wheel rotates the

bar magnets will create a magnetic field between the rotating flywheel and the electro-magnets which

are placed in the plastic housing which covers the aluminium frame of the product. The three electro-

magnets will be controlled by a battery operated 24v motor which is capable of running at 5 different

predetermined voltages between 0v and 24v. By increasing the voltage from the motor to the electro-

magnets the strength of the magnetic field between the bar magnets on the flywheel and the electro-

magnets will also increase. This adjusting of voltages to the electro-magnets therefore results in a

magnetic field of different strength. As the flywheel rotates at a constant speed the resistance can

therefore be increased or decreased as desired.

Mechatronic Control

The product will automatically turn on when the pressure sensor placed in the foot plate is activated.

Once activated the settings can be manually selected by using the control panel situated on the top of

the product. The display will ask for a motor setting. The user will be asked to choose and option be-

tween 1 and 5, these numbers correspond to voltage values which the motor will run at. Once a value

is selected the user will then have to choose a length of time which the exercise will run for. The length

of time can be anything up to five minutes, however anything between 10 seconds and 5 minutes may

be chosen. The user can select their required time by pressing the arrow button on the display, the

time will increase by ten seconds every time the button is pressed. Once selected the user is able to

start their exercise. When the set time has been reached the user will hear a beep to indicate their time

is up. There are also three pre-programmed options to choose from and these are detailed in the flow

charts shown.

Components Required

To achieve the mechatronic control specified some electronic components will be required.

Pressure Sensor—The pressure will measure the force exerted on the surface of the foot plate. Once

an appropriate level is reached, this will indicate that a foot has been placed on the surface and the ma-

chine will automatically turn on.

Piezoelectronic Buzzer—This is a high performance buzzer which employs unimorph piezoelectric ele-

ments which can be incorporated into various circuits. The have an extremely low power consumption

and can be used in this product to provide an audible output. The audible output will alert the user

when the allocated time for the exercise has been reached.

http://www.tdk.co.jp/tefe02/ef532_ps.pdf—accessed 4/1/2012

A 555 timer circuit will also be needed to provide the time detection required to perform the exercises.

Below is the flow chart detailing the programming for the first pre-programmed option available in this

product. Pre-programmed option 1 is the warm-up.

Resistance Mechanism 1

19

60

60?


Below is the flow chart for the walking pre-programmed option.

Below is the flow chart for the cool-down pre-

programmed option.

Resistance Mechanism 2

20

60

60?


Part Sourcing

21

Type B roller chain http://

www.chainreactioncycles.com/

Models.aspx?ModelID=9911

£13.99

Martin 25B15 Type B sprocket http://

www.amazon.com/gp/product/

B0045JJNRS

£6.33

Silverline 431911 Bar Magnets

http://www.amazon.co.uk/dp/

B003TNWC8S

£4.55 for 2

Electro magnet G T040-0 5000-24VDC

http://www1.conrad-uk.com

£46.09 for 3

24 volt battery powered motor

http://www.ebay.co.uk/itm/24-volt-Battery-

Powered-Motor-120-WATT-Motor-E-

Scooter-/

£17.99

Corner bracket (two will be joined together to

make tensioner)

http://www.amazon.co.uk/CORNER-BRACKETS-

25MM-PACK-10/dp/B002A5597C/

£0.99 for 10

Tooth chain sprocket for tensioner http://www.ebay.co.uk/

itm/6mm-04B-Steel-Tooth-Chain-Sprocket-Pilot-Bore-04B-1

-Choose-Teeth-/320771700649?

pt=LH_DefaultDomain_3&var=&hash=item76c0e9f594

£3.65

Westfalia ball rubber band tensioner http://

www2.westfalia.net/shops/tools/transport_aids/

load_securers/tension_ropes_and_belts/411315-

ball_rubber_band_tensioner.htm

£7.99 for 12

Pan head rivet

http://www.ebay.co.uk

£7.50 for 25

Flat head rivet

http://www.ebay.co.uk

£8.50 for 50

Truss head rivet

http://www.rivetsonline.com

£5.56 for 100

Pressure sensor

http://uk.farnell.com

£15.40


The material pricing for the project is based on producing one product. The overall cost for the material

required for this project will decrease if all of this material is ordered in bulk. The length

of time spent on the production line and also the efficiency of the design and the produc-

tion will alter the cost price.

Direct and indirect project costs must also be added to the material costs. These are;

Direct wages—the payment of staff directly involved in the creation of the product,

from design and manufacture to marketing and package design.

Indirect wages—The payment of staff who are indirectly involved in the project, for

example admin staff, contracts, personnel and other such overheads.

Materials for prototypes and development aids

Subcontractors

Capital equipment—computers, machines and other such equipment available

within the factory

Accommodation—the facility rental (if required), design suits, heating, lighting

Travel—required for business meetings, marketing requirements and promotional

aids etc.

Consultancy fees

Indirect material—stationary such as pens, paper printer ink etc.

Costing Considerations

Costs may increase due to unforeseen circumstances or problems so a contingency

plan will be needed for events such as difficulties in estimating, design errors, delays

due to strikes, subcontractor errors, and this contingency allowance can be included in

the initial project cost estimate. Inflation rates, wage rises and rising and falling curren-

cy exchange rates will also affect the project cost. This should also be considered in the

initial estimate.

Having taken everything into account and including contingency costs and other such

costing considerations in my estimate I propose that the project costs price to create

one product should be no more than £400. This price will hopefully be less but for the

reasons explained above it is essential to over estimate this price to avoid financial

problems for the manufacturer later.

Information taken from 2nd year Design and Manufacturing Management lecture

presentation accessed 3/1/2012—http://classes2010-2011.myplace.strath.ac.uk//

file.php/10143/Lecture_wk11.pdf

Product Costing and Bill of Materials

22

Material Supplier Quantity Price

20 x 20 Box Section www.metalmaniauk.com 1 x 1.5m length £7.54

30 x 30 Box Section www.aluminiumwarehouse.co.uk 1 x 2.5m length £9.78

Circular Tube Section www.metalsdepot.com 1 x 12ft length £33.11

3mm Aluminium Sheet www.forwardmetals.co.uk 1 x 0.75m length £10.38

‘V’ Shaped Wheels www.ebay.co.uk 4 £46.76

6mm Aluminium Sheet www.ebay.co.uk 1 x 0.6m length £33.00

Rubber Sheet www.cbsonline.co.uk 1 x 0.7m length £4.80

Type B Chain www.chainreactioncycles.com 1 £13.99

Type B chain Sprocket www.amazon.com 1 £6.33

Bar Magnets www.amazon.com/ 3 £6.83

Electro-Magnets www1.conrad-uk.com 3 £46.09

Battery Powered Motor www.ebay.co.uk 1 £17.99

Small Chain Sprocket www.ebay.co.uk/ 1 £3.65

Small Ball Rubber

Band Tensioner

www2.westfalia.net 1 £0.67

Pan Head Rivet www.ebay.co.uk 9 £2.70

Flat Head Rivet www.ebay.co.uk 2 £0.34

Truss Head Rivet www.rivetsonline.com 16 £0.89

Polythene Chips compare.ebay.co.uk/like/360365610470?

clk_rvr_id=304126012261

1 x 40x24x48 sheet £5.70

Pressure Sensor http://uk.farnell.com 1 £15.40

Raw aluminium www.metals4u.co.uk 1 inch aluminium rod £15.02

Corner Bracket www.amazon.co.uk/ 2 £0.20

Total Material Cost £281.17


CAD Rendering and Views

23



24

Both plastic casing elements have snap fits included to pro-

vide an economical way of joining the elements together to

successfully encase the frame and working elements of the

stepper.

Bosses have been incorporated in this

design to provide secure housing for

the electromagnets and electrical com-

ponents required within the product.

There is a channel to allow wiring to

be secured and hidden and ribs are

placed on the outside walls to help

with manufacturing.

In the base component a large rectangular section has

been cut into the thickness of the material to accom-

modate the frame which supports the functional mech-

anism of the product.

Ribs have been included

to help strengthen the

material around the boss

area to help prevent

warping.

As the pieces are quite long and nar-

row ribs have also been added over

the surface area on the inside face

of the component. This will increase

the strength by increasing the wall

thickness and reduce the amount of

bending which can occur within the

material.

A cut has been place in the surface of the

top component to allow movement of the

wheel driven slider mechanism. This could

provide a potential component collision

point if the length of the hole is insufficient

for the length moved by the mechanism.

Cuts into and

through the surface

of the top component

have been made to

accommodate the

control display unit.

A minimum 0.5° draft angle should be included for manu-

facturing because of the injection moulding process. A

0.5° radius has also been included on all edges on the in-

side surface of the component to enable removal of the

component from the mould,

On the outside surface of the base component a cir-

cular undercut has been included to allow sheet rub-

ber to be cut and placed appropriately. This will sup-

ply the grip necessary to prevent the product from

moving over a slippery surface during use.



25

This picture shows an exploded view of the fly-

wheel assembly. Many of the components here

can be sourced, the bearings, the sprocket and

the bar magnets. The only component which will

need to be specifically produced is the flywheel.

As the flywheel has bosses on the surface to

provide a housing and location tool for the bar

magnets, material has been removed in the form

of holes to try to save weight.

The control display unit requires three

components, the top housing, the PCB

and the backing. These are shown in an

exploded view.

The slide rail mechanism assem-

bly is shown. The plate alumini-

um and the box section required

for this can be sourced but ma-

chining will be necessary.

The wheel-driven sliding mechanism is assembled with the compo-

nents shown. The rivets, ‘v’ shaped wheels, box section, rubber sheet,

aluminium sheet and chain tensioner required can all be sourced from

other resources. Machining will be required on most of these compo-

nents once sourced. The foot rest will be made with a metal casting

process, the raw material required for this will need to be purchased,

consideration to the surface finish of this component will need to be giv-

en as this component is situated on the outside of the product where it

can be touched and easily seen. Bushes have been used between the

‘v’ shaped wheels and the aluminium plate to prevent the plate from

colliding with fastening used in the slide rail assembly as the foot rest

moves forward and backward. The chain tensioner is also there to pre-

vent the chain which will need to be attached to the mechanism from

hitting the wheels during motion.

This is the exploded view of

the plastic connector assem-

bly. This component connects

the chain to the frame via an

elastic rubber band tensioner.

This keeps tension in the foot

rest, returning it to its original

position when no force is being

applied by a user. It also con-

nects the chain to the foot rest

to provide the resistance re-

quired. It does this by using a

small sprocket which can be

sourced.



26

This picture shows the assembly of the final product. The

frame assembly is inserted into the correct place in the base

and the chain is attached. Part of the chain assembly is

shown below. The chain attaches to the sprocket in the fly-

wheel assembly, loops around the small sprocket in the plastic

connector and then attaches to a securely fastened rivet in the

wheel-driven sliding mechanism, passing over the chain ten-

sioner to avoid collision with the drive wheels. The plastic

connector is then attached, via an elastic rubber band tension,

to the front tubular section of the frame. As the flywheel then

rotates a force field between the bar magnets and the electro-

magnets creates the resistance required.

Section view in top

plane.

The main metal frame will be

assembled as shown here be-

fore being place into the plastic

casing designed specifically to

house the frame.

Integrating Studies 3 project

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Transcript of Integrating Studies 3 project