LIM4102 – Design of Mechanical Systems
Prof.: Melvyn Alvarez Vera
Project report
Design and analysis of an automatic tube bender
Student information
ID: 163598
1
Spring 2023
Executive summary
In order to obtain the final tube bender, the following process as mentioned above is planned and carried out
according to the information based on surveys and the needs of the proposed companies. Further analysis it is
presented the following results obtained from the process mentioned:
On the questions related to both options we couldn't see any big difference between both propossals because
both were designed similarly and in a way both options were similar in the making of. Even though there is
not much difference, the second option has better grades overall as it can be seen making it a better option to
consider.
2
Spring 2023
Moving onto the House process (QFD), it is evident that the competence is way better than que proposals
shown but the price is way higher because the quality and the size is more industrial. The proposals are less
expensive but we see a better performance for the second proposal in general and on the needs mentioned
above.
As same as the house method, the multi level method used which results in a better grade for the second
proposal as mentioned.
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Spring 2023
As the previous methods state that the second proposal is better, we develop a cost method where the parts
cost is shown in order to compare which proposal is less expensive and more affordable to carry on and to
have better earnings over-time.
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Spring 2023
Finally, we use the data obtained and create both NPV and a year investment plan which lead us to believe
that the second proposal can have better earnings and produce more logic data which with an investment of
3,804,372 mexican pesos can be obtained within 2.6 years. It is important to say that the plan refers to a 5
year plan which can be traduced to the earrings of more money of the customer.
The Gannt diagram is provided which explains the development time and the process that were followed in
those days in order to achieve the conclusion of the second proposal being the best and the developed in the
project above. The colors represent the process in time and month that were done (taking into account that
some processes were done in the same days).
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Spring 2023
Index
Section
Page
Executive Summary
2-5
Abstract
7
1. Introduction
8-11
2. Identification of needs.
12-18
3. Problem definition
18-20
a) Generation of multiple solutions (functional phase)
21-23
b) Advantage/disadvantage chart or table
23-24
c) Decision-making process
25-28
4. Bill of Materials
29
5. Design of Mechanical Elements
30-46
6. Finite Element Simulation Analysis
47-64
7. Functionality Analysis (Cinematics)
65-67
8. Documentation and Project Support
68-73
References
74-76
Appendix (Technical drawings) and Statistics
77-96
Patent
97-101
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Spring 2023
Abstract
With the introduction of more technologies throughout history, the necessity to automate manual systems is
almost a need in order to facilitate and be more productive while producing in quantity. Applying this in the
area of bending of steel, to be more specific “tube bending” and its increasing demand, it is important to
explore different options and ideas in order to achieve a way that can benefit both the inventors of those
machines, and the companies that use them.
In this article, it is explored two automated proposals compared to two existing ones to discover if
they are a better option for tube bending and for small/medium companies that may purchase those machines.
It is explored from the making of those machines, the cost of production and if its cost-worth it for the
consumer and the seller.
Other systems to analyze the proposals are used such as quality standards, the QDF method, VPN and
the actual analysis of each piece of the structure in order to obtain a precise and objective result.
Later on, it explores the calculations of each of the pieces of the system including its own comparison
with a FEM program called ANSYS which will help us to know if the propositions work and the
arrangement is well calculated to have an exact and precise tube bender. Technical drawings were attached to
further analysis and observation.
The addition of the manual and different patent information is neccesary in order to create a better
and precise product for the customer. The modification of the system itself was done in order to complete the
system proposed.
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Spring 2023
1. Introduction
Steel bending has been one of the most evolving forms of engineering since the nineteenth century. Initially,
the main form of creating bended geometries was with the use of molten steel and later on through the use of
forge to achieve certain curves, angles desired. Moving on to the twenty century, Steel bending was more
frequent but still bended by metal benders by hand.
Later on in the century, three roll bending machines were introduced and this created an interest in
developing more ways to bend steel in many other forms such as sheets, tubes and beams within its
variations. At this point almost every way to bend a steel was manual and all the force within the help of the
machines was put in by a human worker that controlled all the system. [1]
Figure 1. Three roll bending machine. (From: Wammes machinery)
During the last two decades of the twenty century there was a significant increase in demand for bent
steel (curved) so technology continued to grow according to these neccesities. As mentioned, a variety of
steel can be bent into many angles, forms nowadays but we are going to be focused particulary on the
bending of hollow tubes due this shape can have enormous ways of being used in the industry.
Another important part of the bending machine system as stated above is that the speed, price, quality
and force used to bend the tube is that almost entirely depends on the worker using the machine as a result
that the machine is entirely manually controled. What is being used now, are systems that almost are
automated and only being monitored by a single person which is a significant increase in all of the
qualifications mentioned.
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Spring 2023
There are already existing in the industry a wide variety of automated systems which vary in size,
price, quality and speed according to the needs of the company that is using them. There are many types of
these systems where everything is connected and are way more productive to the manual ones as a matter of
fact.
Figure 2. Automatic tube bending machine. (From: SMC)
As an example, the picture above is presented as an example to see how the basics of a tube bending
machine works, as it is possible to see, the tube is being pressed as desired into the die system that later on
the system bends depending on the angle needed. All of these processes, as seen, are only being monitored
because the machines and the motors do all the work that in the past many workers needed to do.
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Spring 2023
To be An engineer:
As any other professional area, engineering requires some of the highest standard’s behaviors within the
principles of ethical conduct as is stated in the National Society of Professional Engineers (NSPE). Since
May 1935 we can find the first form of the American Engineer code but in the form of a “suggestion for a
membership” (NSPE) instead of the way of life that nowadays, a professional engineer should live. [2]
The Code of Ethics is all about the six main professional duties which are divided into more complex
and simple standards that are listed below and later explained with more detail to have a better understanding
of them:
● “Hold paramount the safety, health, and welfare of the public.
● Perform services only in areas of their competence.
● Issue public statements only in an objective and truthful manner.
● Act for each employer or client as faithful agents or trustees.
● Avoid deceptive acts.” [3]
Conduct themselves honorably, responsibly, ethically, and lawfully to enhance the honor, reputation,
and usefulness of the profession.
The order of the canons does not reflect if one of them is more relevant than the other, all of the
subjects matter the same in the same level and should be considered equally in any situation if applicable.
Starting off with the first one, it all relies on one as an engineer should have the judgment and the
criteria to know and say if a certain circumstance is appropriate, valid and it's all within the right part of the
law. This section also contains information about how standards and honesty should be applicable to the
same circumstances always without any doubt.
Moving on to the next area, it is a common sense of basic ethics in which a professional engineer
should only take a project and labor on the area that it is specialized on and no other subject. This, to ensure
the responsibility and the coordination of an entire project within its quality itself without altering the final
product.
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Spring 2023
As the next subject, objectivity and subjectivity are meant to be separated pointing towards the truth
in any manner. No criticism shall be issued, and every information presented needs to be founded within real
and objective information.
Moving on to the final point, it is a mixture of both honesty and the objectiveness of a previous
statement where any engineer should not misrepresent information of anything that he/she does not process
starting from its professional abilities.
Of course there is also on the list a section called Professional Obligations which complements the
aptitudes of a professional engineer referring to how they should acknowledge their errors without altering
the reality (lie). Also, it is added that they need to be encouraged to see and participate for the wellness of the
community.
To finish the brief explanation of the last code lines, it is important to acknowledge that engineers
shall not affect in any way other professionals and accept their responsibility in their own activities including
their respective credit in it without exaggerating or false accusations. Now, after having a brief explanation of
what to do and know as an engineer, it is important to apply this knowledge and not just read it.
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Spring 2023
2. Identification of needs.
As it is known, tubes have a wide variety of applications in many ways and many daily life activities that we
may not be aware of at first but, with a little bit of time, we may see that we are surrounded by bent tubes
almost everywhere we go. One simple example is when we are in the bus, bathroom, kitchen, we can find a
lot of different tubes from many materials that have different uses but if we look deeply into the tubes, they
are bent.
Figure 3. Example of a tube on a daily basis. (From: Bender)
Companies tend to work with different varieties of tube materials, size and thickness in order to
provide more solutions and to have a better market for their sales. When we talk about these companies we
talk about gigantic ones that can purchase and handle such machines which is very complicated for small and
medium businesses that want to bend tubes for personal and commercial use.
In this case, we want to provide a machine that will focus on speed and quality instead of variety, of
course there will be a range of variety in size and material to work with but we are not going to focus on any
other characteristic in order to provide an affordable, easy to use and efficient machine for those business that
cannot work with bigger and expensive machines.
The steels that are going to be used as an example for our machine to work with are the next three
common steels used in the industry which are 316 stainless steel 304 stainless steel and 201 stainless steel
and in the shape of hollow tubes.
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Spring 2023
Figure 4. Example of a hollow tube. (From: Tubos)
As mentioned before, the project is aimed towards small and medium businesses that on average is
calculated that they will bend 50 pieces maximum a day so having this in mind and that the machine is
“fully” automated and it will be easy to operate, only one worker will be needed checking the machine while
functioning. The look and appearance of the machine will not be a main thing but it is going to be intended to
look decent enough to the human eye.
Why are we focusing on small/medium size businesses? For the simple reason that big ones have
more facilities and many resources that the others dont, so in a way of providing those tools we will develop
more options for them to choose from.
Moving on to the precision of the bending, it is very important to achieve it because the people that
will work with will rely on the quality of the final product. Standar ISO9001 will be taken as adequate
measures for all of the process and the products that will be taken into account.
Starting the analysis of both of our proposals, it was conducted of what is known as a quality function
development (QFD) which is a technique to obtain a diagram with all the needs that must meet a project, in
this case, the proposal of our two machines. This is a way to optimize the results and the product itself to
have a better quality for the needs of the buyer.
The QFD method was done by asking 10 people what they would consider more important on a scale
1 through 5, 1 being less important and 5 more important. The parameters used are shown below and will be
considered to be epplied to the final product.
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Spring 2023
This are the results for the first proposal shown:
Table 1. Results for the first proposal.
Table 2. Results for the second proposal.
As mentioned before, we can see how the people interested follows an score of 3 and above for
almost every characteristic mentioned for the machines shown, what we can see is that even though the
results are almost the same, we can see that the second proposal have a better average score for all of the
parameters compared to the first one.
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Spring 2023
Table 3. Results for the second proposal in the Pie chart.
Table 4. Results for the second proposal in the bar chart.
In addition to the first graph, a pie and a bar graph are added in order to compare the information being
shown. First we can see and say that the pie graph is no way to plug in this information because of the almost
the same distribution in the values making it impossible to contrast any sort of possible information.
Finally, the bar and the candle graph can be more accurate and easy to use in order to compare
information for the case of the data such as the easy to use value being the breathes and the variety of tubes
the least important. With this we can see how the bar graph is way more accurate to use in this project.
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Spring 2023
Now that the test was concluded, we can continue to finally use the data obtained to create our QFD model as
mentioned. It is important to say that according to the necessities obtained, we will be focused on all the
parameters equally to have a balance between them and not to have one being better than the other.
For the reference, Competence #1 and #2 are as mentioned in reference [15] and [16]
Table 5. QFD model for both proposals.
Now we have a point at where to start in the making and planning of the machines that are going to be
built, as looking at the data, the durability and the price of the machine are both the most important
characteristics to have in mind. The time of production and the force required for the work also will be
taken into account, along with the number of parts and the angle of bending but will not be as
important as the other ones mentioned.
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Spring 2023
For the HOW section:
● Angle of the bend: The variety of angles at which the machine can work, from 0 to 180°
depending on the design/arrangement of the model.
● Time of production: How fast can the machine bend a tube from start to finish.
● Price: The final price of the machine
● Durability: How long will the machine last without having to repair it or before it starts to fail.
● Force: The needed force (from the motor and the system itself) to bend a tube.
● Number of parts: The final parts that the machine will have in total.
For the WHAT section:
● Machine life: As similar as the Durability, it refers to the life that the machine will last before
having problems.
● Speed of work: How fast can a tube be bent from start to finish.
● Precision: How exact the final angle will be produced (can have margin of error).
● Maintenance: Refers if the maintenance of the machine will be easy or not, including the
frequency of it.
● Easy to operate: As the name itself, depends if anyone or many people will use it easely.
● Quality of the machine: Is how the final product will look with its finishing surfaces.
● Safety: Refers if necessary certain or extra protection compared to a regular bending machine.
● Cost effectiveness: If the price of the machine compensates the earnings of the company
through time.
● Variety of tubes: The variety of length of the tube, thickness, types of tubes and materials that
the machine will be capable of working with.
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Spring 2023
We will use a Multi-weighted scoring model in order to have a better impact on understanding of our score
system. It is important to have in mind that the maximum score possible would be 5.
Criteria
Machin
e life
Speed
of
work
Precisi
on
Maint
enanc
e
Price of
machin
e
Easy to
operat
e
Quality of
the
machine
Safet
y
Cost
effecti
veness
Variety
of tubes
Total
Weight
4
3
4
3
4
4
5
5
5
2
39
Proposal
1
2
2
3
2
3
3
4
4
3
2
28
Proposal
2
4
3
4
5
4
4
5
4
4
2
30
Compete
nce 1
5
4
5
4
3
4
5
5
4
4
37
Compete
nce 2
5
4
5
4
3
4
5
5
3
4
36
Table 6. Multi-Weighted score model.
3. Problem definition.
Nowadays, It is a fact that there are tons of machines that bend tubes all around all kinds of companies that
come from a considerable amount of years. Indeed, they have been evolving but the limitations when it
comes to real life start to show up starting as mentioned before, the price itself of the machine.
Most of the time, small and medium companies tend to go directly to bigger ones because there is no
need for them to buy an expensive machine in order to bend certain tubes for their products. What is intended
is to produce and promote a useful machine that will allow those businesses with their bends.
The second issue that is found refers to the material's hardness that not all the machines can work
with, before automation, most of the bends were directly possible to the force applied by the worker. Now
with the introduction of motors, we can secure a more precise and clean bend.
Another problem faced, the angles at which we will allow our customers to bend their tubes, most of
the bends are below the 90 degrees but most of the times for smallest businesses, require more specific angles
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Spring 2023
due the experimentation of the products that they produce. The angles that will be taken into consideration
are at a starting point of 0° and a possible angle bending of 180°.
In addition to the information above, it is important to have in consideration the yield point of the
tubes that are going to me bent to later on take that value in consideration for our machine to properly work.
The steels mentioned again, are 316 stainless steel 304 stainless steel and 201 stainless steel which have a
yield strength between 193 to 215 Mpa.
Standards-Norms:
As an addition to the problems mentioned, we must focus and build our design around certain standards that
are necessary to follow as engineers in order to obtain a better quality product and to stay in contact with the
real world needs and measures/rules.
Standards considering proposed solutions
Standard Number
Description
PFI - ES24 [8]
“This standard covers methods, process requirements, tolerances and
acceptance criteria for shop fabricated pipe bends.”
SAE - AS130 [9]
Refers to standard dimensions related to tubes that can be used on this
type of equipment such as the radius.
ANSI B11 [10]
It is necessary to know about the bending machine safety requirements
that include rollers, dies.
ISO - 13857 [11]
This defines the safety parameters but refers to the machinery in
general in order to protect the operator and limiting the danger zones.
ASME - Y14.5 [12]
ASTM A269 [13]
Tolerances and dimensions at the moment to interpret the designs.
Refers to the tubes made of stainless steels and its standard parameters.
Table 7. Standards used according to our model.
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Spring 2023
Constraints according to the standards mentioned:
● Material used: As stated below, stainless steels would be the main source of work with its according
standard such as de ASTM A269 with a diameter permission from 2 cm to 4 cm in order to maintain a
fair price/size of the proposed machines.
● Properties of the working material: All of the properties of the working steels should remain intact
and not be affected negatively as the ASTM A269 states so it is important to have that data in mind to
not reach the ultimate stress of the steel.
● Size of the system: As intended, the system is aimed at small and medium size businesses so we will
be focusing on producing one that has a decent/easy to move system compared to the enormous ones
that we see on bigger companies.
● Safety zones: As the ANSI B11 standard follows, it is important to state and clarify all the working
zones within its limits for the operator in order to maintain the safety of him/her at every moment.
● Number of tubes per work cycle: As its name states, to maintain the design and the order of the
system, it is intended to work one tube per bend also to keep it simple.
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Spring 2023
Synthesis
a) Generation of multiple solutions (functional phase)
For both proposals shown, four different angles of the system will be shown followed by a brief explanation
of the system with some lines of reference for the movement itself.
First proposal system
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Spring 2023
As the system itself, it consists of two connected main gears where the small one is connected to a
motor which makes it rotate and finally giving the move of the bigger one with as we can see, the central die
stays in its place while the second rotates within the same movement of the big gear.
As easy as that and the larger circle with holes in it with the automated arm which only serves as a
stop sign for the angle required for the system and it will change places depending on the angle with a
rail-system on the bottom. The tube can be introduced both ways of the table and is intended to be placed
near a wall due the form of the table legs to compensate for the weight of the parts.
Second proposal system
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Spring 2023
The main difference on the first proposal is that this system does not contain gears but instead, it uses
a single motor where with the addition of the rail-system of the outer side of the platform, it will bend the
tube that is introduced by only one place as seen on the hole. It contains a similar system of holes which will
help for a second sensor-motor to stop and introduce a pin in order to prevent any further movement.
This system, as already mentioned, is intended to use it in only one way and can be placed anywhere
near the center or anywhere where the tube wont hit any wall due its circular geometry of work. It is planned
to be a simple system where the arm rotating stops at a desired angle.
b) Advantage/disadvantage chart or table
Advantages and disadvantages of tube bending machines
Advantage
Disadvantage
Proposal # 1
Due its design, it can be placed near to
walls to save more space in the building.
If needed, it could not be placed
anywhere else so it is restricted only to
those places next to a wall.
Proposal # 2
Can be placed anywhere near the middle
of the building.
If it is located close to a wall, the system
may not work properly.
Proposal # 1
Its original design may bring a great
looking acquisition to the place.
Due to its originality, some pieces of the
mechanism will be hard to find to
replace.
Proposal # 2
A wide variety of angles for the bend
giving more choices to the customer.
The product itself could increase in price
due its sensor/force applied system.
Proposal # 1
Could be used in two different ways, left
or right.
This same advantage can be a
disadvantage if it is required to use a
larger tube.
Proposal # 2 Due to its arrangement, it can be faster on
the work and only be focused to one tube
per cycle.
Proposal # 1
Very easy to use for the operator due its
self-explanatory design.
Can only be used in one way.
May-be a little bit dangerous if not used
properly because the movement of the
gear or the system itself is exposed.
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Spring 2023
Proposal # 2 Same easy use for the operator as the first
proposal.
Can be a little hard to transport due its
circular tall shape but once placed can
save a lot of space.
Proposal # 1
Easy to assemble due its reduced number
of pieces.
The main pieces can be expensive and
can weigh a lot.
Proposal # 2
A great addition to a company due its
great ergonomic looking design
(esthetic).
May look like a hard machine to
assemble at first.
Table 8. Advantages and Disadvantages chart for both proposals.
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Spring 2023
c) Decision-making process
Depending on the requirements established by the consumer, the final proposal will be chosen as the way to
go on the project and which will be the best fit to its necessities. It is important to move to the analytical part
where the price, budget and possible earnings planned will be taken into account in order to have a final
decision for the machine.
Cost of the first proposal
Pieces #
Material name
Cost $
1
Aluminum table
5,000
1
Aluminum small gear [14]
1,200
1
Aluminum big gear [14]
2,000
1
Tube for the small gear
150
2
Die
1,600
Total
10,400
Table 9. Estimated cost for the first proposal in Mexican pesos.
Cost of the second proposal
Pieces #
Material name
Cost $
1
Aluminum Table (Total)
9,962.4
2
Die
9,000
6
Ptr profiles
3,926.2
1
Motor
3,024
7
Plate + pin + bar + bolts/nuts
1,302.5
6
Ball bearings
488
1
Shaft
4,000
Total
31,703.1
Table 10. Estimated cost for the second proposal in Mexican pesos.
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Spring 2023
Prices for some things may vary such as the tables and the gears because they are very specialized
and considered to be homemade by buying it and by the means of a milling machine. It is just an
aproximation and it will be reconsidered such as the initial price of the aluminum base to work with.
Now, for the final part of the cost estimation, a Net Present Value (NPV) model based on 2 machines
sold monthly because even though we want to sell tons of machines, the focus of the project is based on more
specific cases which will sell less that commercial models (important to say that the machine will be sold at
59,999$ mexican pesos).
NPV for the first proposal
Project 1
Year 0
Required
Rate
Year 1
Year 2
Year 3
Year 4
Year 5
Total
20%
Outputs
-100,000
-100,000
Inflows
249,600
249,600
249,600
249,600
249,600
1,740,000
Net
Inflows
249,600
249,600
249,600
249,600
249,600
1,640,000
NPV
646,457 $
Table 11. NPV values for the first proposal in Mexican pesos.
NPV for the second proposal
Project 1
Required
Rate
Outputs
Year 0
Year 1
Year 2
Year 3
Year 4
Year 5
Total
20
%
-3,804,372
-3,804,372
Inflows
1,439,976 1,439,976 1,439,976 1,439,976 1,439,976
7,199,880
Net
Inflows
1,439,976 1,439,976 1,439,976 1,439,976 1,439,976
3,395,508
NPV
502,038 $
Table 12. NPV values for the second proposal in Mexican pesos.
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Spring 2023
After all that data, we calculate the Payback period and the rate of return to see if the value of the product and
the business will be retailable.
Table 13. NPV values for the second proposal in Mexican pesos.
It is important to clarify that for the NPV models positive values against negative ones mean that it is
a good project. Taking this into account we see that both proposals are a good investment but we will select
proposal number 2 because we consider that the earnings are way better than the first one.
Finally, a Gantt Diagram is provided to show the organization and the plantation of the project from
the day it was propossed till the last day it is being worked on. This is only used as an agenda and to have an
itinerary for future reference.
Table 14. Gantt Diagram for achieving progress.
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Spring 2023
d) Considering a real situation
The considered situation, as already stated above, is considered to be a business that will bend an average of
50 tubes per week or even more which is going to be closer to an independent business that needs a very
good variation of angles for those tubes. The size of the tubes will not matter more or less because the small
and medium businesses rarely work with big tubes so we will focus on angle variation and speed of
production.
As the final selection, the second proposal will be chosen due the previous statement which provides
a more variety of angles for bending and with the combination of the ergonomic design, the faster speed of
production compared to the first proposal, it will be a better acquisition for those businesses and probably
will bring more sales. The materials that will be worked with are already stated as those three stainless steel
such as the diameters previously mentioned, the machines will be built according to the force needed to bend
those tubes.
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Spring 2023
4. Bill of Materials
Bills of Materials
Product: Semi - Automatic – Tube bending machine.
Date: 06/03/2023
Assembly: XXXXXX
Item
#
Part #
Qty
Name
Material
Source
1
JR011
1
MOTOR 1.5 HP MONOFASICO 3600
steel
Amazon
RPM JR011
2
N.20-CON-VENT
1
Shaft Azteca N.20
Stainless Steel
Mercado Libre
3
155
6
Hex bolts, Stainless steel 18-8,
3-18 SS
Bolt Depot
3-18 SS
Bolt Depot
6061 Al
Ming Tai Ala
May vary (Steel)
Construrama
3/8"-16 x 1/2"
2556
4
6
Hex nylon insert lock nuts, Stainless
steel 18-8, 3/8"-16
5
6061 Aluminum Alloy
plate
4
6
CMA0205010020
5
7
0758226998020
1
Aluminum plate for the supports
H13 Aluminum
MUKCHAP
8
H13-Die for Welding
6
H13/SKD11/D2 Roller Mold/Die for
Welding Pipe Tube Mill Roll
H13 steel
SENBO
2
Ball bearing
Steel
Amazon
9
Balero 6205 Rz
6061 Aluminum Alloy Plate for
supports and Table Plates
Ptr 1 1/2" X 1 1/2" Azul, Pieza for
table legs
10
X60028
3
smaller Pin
Steel
Globe
11
30CrMoA/4130/SCM430/
1.7218
1
4130 1´ Bar for Support pin
4130
Alibaba
12
1100 3003 5052 5754
2
Custom 6061 plate for holders
AL 6061
Hongtu
Prepared By: Carlos INC
Company:
Bending UDLAP
Checked by: Carlos INC
Page: 1/1
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Spring 2023
5. Design of Mechanical Elements
Motor (1)
The following calculations were necessary in order to determine critical and important parameters that the
designed tube bender will need to have to secure a precise and safe final product. First, we will focus on the
force needed to bend our tubes, it is stated that the maximum diameter that the bender will support is 4 cm so
we will take into consideration that data.
Figure 5. How the system works reference. (From: https://www.wermac.org/specials/pipe_bending.html)
Now, we will obtain the basic data from our tubes and the material itself as it follows being the AIS 304 and
the properties of the tube from reference (17).
Value/Units
Material
AISI 304 (Stainless steel)
External diameter
4 cm or 0.04 m
Internal diameter
3.76 cm or 0.0376 m
𝜎 Yield stress
215 mPa
Table 13. Main data to calculate bending force
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Now we proceed to use the formula for the bending moment obtained from (18) to obtain the value needed.
𝜎=
𝑀*𝑐
𝐼
(1)
𝜎𝑦*𝐼
𝑐
(2)
we solve for the Bending moment M so we obtain:
𝑀=
M= bending moment
C= distance from the neutral axis
I= moment of inertia from the cross section
𝜎= normal stress
Now we need to define more of these values such as the moment of inertia which is calculated as it follows
the next process:
𝐼=
𝛑
4
4
4
(𝐷𝑒𝑥𝑡 − 𝐷𝑖𝑛𝑡 ) =
π
4
((
0.04 4
)
2
−(
0.0376 4
))
2
4
= 2. 75𝑒 − 8 𝑚 (3)
Now we calculate the distance from the neutral axis c:
𝑐=
𝐷𝑒𝑥𝑡
2
=
0.04
2
= 0. 02𝑚 (4)
So with all the data we can plug it in and obtain our bending moment which results in:
𝑀=
𝜎𝑦*𝐼
𝑐
=
215 * 2.75𝑒−8
0.02
= 295. 6 𝑁 * 𝑚
Now with the force required calculated (Torque) we choose a motor that will generate that moment force
(mentioned in the bill of materials) which results in the perfect bending of the piece. Knowing the force
needed to bend the tube, We know that the formula to calculate the Necessary HP of the motor is equal to the
Torque obtained times the rpm divided by 5252 which by looking at motors, the JR011 1.5 HP is the perfect
fit for our system.
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Shaft (2)
Moving on to the shaft that is directly connected to the motor, we proceed to use the formulas obtained from
[19] we know that to calculate the shaft diameter we can apply the following formula:
𝐷 =
3
16 * 𝑇
π * τ𝑚𝑎𝑥
(5)
Where:
D is the shaft diameter (in)
T is the torque used in lb-ft
ꞇmax is the maximum shear stress in Psi
By applying the formula, we get a shaft diameter of .4 inches but to overestimate our calculations and for
safety reasons, we propose a shaft diameter of 1 inch (2.54 cm) which will not increase the price and the
production cost that much. Now with the proper diameter, we proceed to design our shaft based on the results
previously stated which in order to maintain minimum shaft damage, we will keep the shafts as small as
possible meaning that we will place the motor as close as the surface as we can see in the following design
proposed (also important to state that the shaft was home-made with the material proposed in the item list but
it was taken that example as a price reference).
Figure 6. Motor place and shaft reference.
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Bolt (3)
Moving on, we proceed to calculate the bolts and screws that are going to be used to hold the motor which, as
we know, weighs 15 kg (rounded to 25 kg to overestimate the system and including the weight of the shaft
which is connected to the motor itself). From reference [22] we will use the following formulas to calculate
the bolt parameters.
σ=
𝐹
𝐴
(6)
𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑏𝑜𝑙𝑡 =
𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 = 𝐶
𝑛=
𝑃
𝐷𝑛
σ
6
(7)
+ 𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑏𝑜𝑙𝑡 (8)
𝑆𝑝
𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑
(9)
Where:
σ is Stress
F is Bending force being applied
A is the area of tensile stress
C is the Stiffness constant
P is the Thread pitches
Dn is the major nominal diameter
n is the Safety factor
Finally, Sp refers to the minimum test resistance for 1 inch
We know for a fact of the motor specifications, that the screws and bolts diameter can measure a maximum
of .4 inches due the holes in the base that can be attached. With this in mind, we will now state again, that the
screw length .5 is due the configuration of the system which is the motor attached to a 1 cm plate.
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Note that the screw length can vary and can be bigger, this is just a clarification to have in mind so we
will focus on the screw diameter in order to get the other calculations needed.
We took a bolt that have the following characteristics:
Figure 7. Hex bolt used with length and diameter. From:
https://www.boltdepot.com/Hex_bolts_Stainless_steel_18-8_3_8-16.aspx
Now, we will calculate its respective values in order to compare them later on on the finite element analysis:
σ=
245.25
0.1599
= 1, 534 𝑁/𝑚2 = 1. 53 kPa (10)
𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑏𝑜𝑙𝑡 =
1.53
4 (𝑏𝑜𝑙𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚)
𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑓𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 =. 32
𝑛=
1
0.375
85
1.23
=. 38 𝑘𝑃𝑎 (11)
+ 0. 38 = 1. 23 𝑘𝑃𝑎 (12)
= 69 (9)
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After calculating our first information referring to the motor being used, we know its specifications and now
it's the time to confirm if the structure proposed it's going to hold together with the weight of the motor itself.
We will weld the 6061 Aluminum alloy plate that measures 30 cm of each side to the main structure profiles
in order to hold the motor.
Nut (4)
For the NUT calculation it will not be calculated anything and it will use the parameters already mentioned
because it is considered that the nut is not a critical element of the system. In the same catalog (Bolt Depot) is
a link where you can choose which nut is appropriate for the chosen bolt so the following one was used
moving to the following part
Figure 8. Hex nut used with length and diameter. From: https://www.boltdepot.com/Product-Details.aspx?product=2556
Motor support (5)
Having the motor attached to the Table support which is welded to the PTR profiles of the table as seen on
figure 6. It is necessary to calculate if the weld holds the weight of the system so in order to confirm if the
proposition is correct we calculate the following (Note: The main table structure is home made with the items
listed above and noted as which part is which):
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Figure 9. Torsional list information regarding the weld that is going to be used [ 21 ]
Moving onto the calculations, it is important to state that the Tensile Strength (Sut) and the Yield Strength
(Sy) (kpsi) of our plate is 42 and 35 respectively. Also, it is important to clarify that the electrode that will be
used is E7010 to weld our structure.
Now we proceed to state our variables already knowing that d will equal to 30 cm or 11.8 in and b
being the distance that will separate the weld is 40 cm or 15.7 in which is the information that we are going
to be using through the system. We follow the next equations in order to calculate the permissible loads that
can be applied.
ꞇ´ = ꞇ´𝑦 =
𝑉
𝐴
=
𝐹
1.414(5/16)(11.8)
= 0. 19 𝐹 𝑘𝑃𝑠𝑖 (10) [21]
Where ꞇ´ represents the allowable shear stress being applied to the system while V(F) is the allowable force
applied and the one that we are going to be using while the data below A is information regarding the type of
weld that is being used. Moving onto the Ju value we know that:
2
𝐽𝑢 =
2
𝑑(3𝑏 +𝑑 )
6
2
=
2
11.8(3*15.7 +11.8 )
6
= 1, 728 𝑖𝑛3 (11) [21]
𝐽 = 0. 707 * ℎ * 𝐽𝑢 = 0. 707 * (5/16) * (1, 728) = 382 𝑖𝑛4 (12) [21]
And then, we proceed to calculate:
ꞇ´´ = ꞇ´´𝑥 = ꞇ´´𝑦 =
𝑀𝑟
𝐽
=
5(𝐹)(5.9)
382
= 0. 077 𝐹 𝑘𝑃𝑠𝑖 (13) [21]
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Maximum shear:
2
2
ꞇ𝑚𝑎𝑥 = 𝐹 ꞇ´´𝑥 + (ꞇ´𝑦 + ꞇ´´𝑦) (14) [21]
The result being 0.54 F kPsi so we finally calculate our data for the min ꞇallowable which we will use the
data for the E7010 electrode from the following table and for the data mentioned above for the Sut and Suy
of the aluminum 6061 alloy which results in the following data:
● 6061 Alloy ꞇall = min[0.3(Sut), 0.4(Sy)] = min[0.3(42), 0.4(35)] = min[(12.6),(14)]
● E37010 ꞇall = min[[0.3(Sut), 0.4(Sy)] = min[0.3(70), 0.4(57)] = min[(21),(22.8)]
Finally, knowing that the min will be 12.6 and the ꞇmax is 0.54 we will use the following formula in order to
calculate the allowable load:
𝐹=
ꞇ𝑎𝑙𝑙
0.54
=
12.6
0.54
= 23. 3 𝐾𝑖𝑝𝑠 (15) [21]
Meaning that our allowable stress that can be applied is 23.3 kips. Just to have in mind, 1 kip is equal to 453
Kg and the motor that will be used is 15 kg. That is way too low for the force that the arrangement of the
system can hold so we can conclude that it is possible to use it.
Table plates and PTR profiles (legs) (5)
As it can be seen, there are two extra plates that are welded to the main (bigger) plate which is the one
holding the system itself. It is also important to know that the same method of welding as the Motor support
is used, being the E7010 due the same material properties and usage.
Because of the system structure those welds does not represent a critical part of the system it will not
me calculated but instead the legs attached are also welded so we follow the same process as the table
support in order to see if the system will resist the force (weight) of the system before tearing apart in the
process.
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It is used the following formula:
Figure 10. Torsional list information regarding the weld that is applied to the legs [ 21 ]
Tensile Strength (Sut) and the Yield Strength (Sy) (kpsi) of our plate is 42 and 35 respectively. The values
for the electrode are already stated above so there is no necessity to repeat that information. Note that b and d
value is 1.5 inches due the squared form of the profile selected.
ꞇ´ = ꞇ´𝑦 =
𝑉
𝐴
=
𝐹
1.414(5/16)(3)
= 0. 75 𝐹 𝑘𝑝𝑠𝑖 (10) [21]
Again, we move tu Ju value knowing that:
3
𝐽𝑢 =
(𝑏+𝑑)
6
3
=
(1.5+1.5)
6
= 4. 5 𝑖𝑛3 (11) [21]
𝐽 = 0. 707 * ℎ * 𝐽𝑢 = 0. 707 * (5/16) * (4. 5) = 1. 08 𝑖𝑛4 (12) [21]
And then, we proceed to calculate:
ꞇ´´ = ꞇ´´𝑥 = ꞇ´´𝑦 =
𝑀𝑟
𝐽
=
7.4(𝐹)(7.4)
1.08
= 50. 7 𝐹 𝑘𝑃𝑠𝑖 (13) [21]
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Maximum shear:
2
2
ꞇ𝑚𝑎𝑥 = 𝐹 ꞇ´´𝑥 + (ꞇ´𝑦 + ꞇ´´𝑦) (14) [21]
Which gives us a result of 72.23 Kpis and because the material and the electrode is the same mentioned
before, we know for the calculations that the min ꞇall from the 6061 Alloy ꞇall is equal to 12.6 and it is
going to be used that value. Finally, having these values lead us to the allowable load which is:
𝐹=
ꞇ𝑎𝑙𝑙
72.23
=
12.6
72.23
= 0. 175 𝐾𝑖𝑝𝑠 (15) [21]
As first, it may look that the 80kg load is too little for the leg but it it's important to keep in mind that we are
talking of 4 legs of the system that distribute the force evenly due its location meaning that in general the legs
can hold up to 320 kg which for the system itself is enough including the tube that it will be adding more
weight to the system (can even hold the system and a extra person just in case).
Support Discs (7)
For this particular case and as it may be seen in the first appendix attachment, this two discs of 0.5 cm each
(big enough) are used to relieve some part of the weight of the system to the motor and to move it to the main
table which can hold with no problem better weight that the motor itself only holding the shaft itself.
The following figure provides a better explanation and a visualization of how these discs can help
reduce the weight of the system.
Figure 11. Support discs function reference.
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As it can be seen, due their bigger size according to the hole in the center of the table, it leads to the shaft to
pass through it and the disc will hold on to the table holding the weight of the other parts of the system.
Another thing to clarify is that we are talking about friction (both discs will hold the weight but they will
probably spin between them and the table so we need to use some type of lubricant for both steels.
Before mentioning the lubricant, calculations were not made for the disc even though one of them is welded
to the table and to the other is free to move because all of the discs area is supported by the table and it is not
possible for them to bend to any direction. Even though they can break, we are talking about 5 cm thick
strong metal discs so with the low weight that the system uses it is certain that they will last enough.
Figure 12. Lithium grease as a reference. From:
https://www.globalsources.com/Lithium-grease/Lithium-grease-1192339986p.htm.
The lubricant is not mentioned in the planes or calculations because the information is based on facts.
According to various references, lithium grease is one of the best lubricants in the industry for metal-metal
friction or frictionless systems in order to move smoothly. Now that it is stated that the lubricant and the
discs' purpose and design, we proceed with the next part.
Dies (8)
As it is known, two dies will be used and will be the same exact model for both cases. Looking at the system,
the dies will be used to bend the tubes, one being the bender and one the support that will remain in its own
place only spinning in its edge.
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First, it is necessary to know the yield stress of our material which is 1640 mPa and we proceed to calculate
the value of 𝝈 obtained from the following formula (used for compression due or the die configuration) [23]:
σ=
𝐹
𝐴
(16)
Where:
F is the force being applied
A is the area
We already know all of the values proposed such as 𝝈 being the yield stress mentioned, and the force we
previously got was about 295.6 Nm but in order to get the total force, we divide this value by the distance
between the bending die and the center of the tube which is 0.082 m which lead us to a force of 3,605 N
(increased to 200,000 N to overestimate the system). Now for the area, we already know the width (which is
proposed) of 16 cm or 0.16 m.
Then, the following equation results as:
1. 65𝑒 + 9 =
200,000
0.16 * 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
(16.1)
So it is solved for the thickness to know the minimum required for the system and lead us to 0.0007 m or .7
millimeters in order to maintain a strong die meaning that it is needed so little die to hold the bending but
even though the calculations, it is almost impossible to have a die with a thickness of .7mm so everything
above that is a good choice.
Note that the Die thickness proposed is about 5 cm which is the necessary maximum for the tubes
proposed before and for many other tubes below that which also increases the visual presentation of the final
product with a bigger die instead of a tiny or even a miniature one.
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Bearings (9)
Now with the respective motor, respective shaft. We know that the system will require at least four
bearings in order for the system to function properly so we now calculate the ones needed according to the
data that we already have which follows as obtaining the loads that will be applied on the bearings and then
selecting the one that match our needs in inner diameter and force [20]:
𝐹𝑟 =
𝑇
𝑑/2
(17)
Where d is the diameter of the shaft (m) and as we know, T is the torque applied in the system which solves
for Fr referring to the radial force. By applying the data we obtain that our Radial force is 23,307 N or
23.3kN. Moving onto the de axial force, we know that the axial force is referring to the force being applied
perpendicular to the bearing as we can see in the following picture.
Figure 13. Loads in a Bearing from: https://www.bearingtips.com/how-do-i-determine-the-loads-on-a-bearing/
Now with that in mind, we know that the force applied perpendicular to our system is approximately in the
axial direction is about 8.4 Kg but again, we will overestimate our system in order to maintain certain
security that it will hold the force being applied with no problem and being taken as 10 kg. Now with that in
mind we multiply that value times the value of the gravity to get the force in N which results in 98.1 N.
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Now we only need to find a Bearing with the data that we have from the catalog SKF which will
provide some of the values that are missing in order to see if the bearing can handle the force applied.
Knowing the Inner diameter of one inch or 25.4 mm we look for that measure and try some values to see
which one will stay still for the forces applied and the arrangement of the system.
Figure 14. SKF Bearing catalog examples and selecting from: skf.com/go/17000
Now that we have the reference values, we first will get the Equivalent Dynamic Load (P) from the following
formula:
𝑃 = 𝑋 * 𝐹𝑟 + 𝑌 * 𝐹𝑎 (13) [20]
Where P is the dynamic load, FR and Fa the radial and axial load respectively and X and Y are load factors
specific to the bearing type which can be found in the manufacturer´s catalogs. For this instance the X and Y
factors are going to be 0.56 and 1.45 respectively due the value calculated of e being 0.33 obtained from
tables of the catalog itself as mentioned above.
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From the formula above we get that P value is about 13.19 kN which now we use it to calculate the
Basic Dynamic Load Rating to select our final bearing. It is also important to have in mind that the basic
rating life factor that is going to be used is an average of 1,000,000 which leads us to the following formula
of.
𝐶=
𝑃
𝐿(𝑙𝑖𝑓𝑒)
=
3
13190
1,000,000
= 0. 24 𝑘𝑁 (14) [21]
And with that information, looking at the bearing catalog we obtain that the bearings as seen on figure 8, we
get to the conclusion that the bearings that will be used are going to be:
Figure 15. SKF final selection of the Bearing from: skf.com/go/17000
Pins and Die´s holders (10,11,12)
For the final part of the calculations, the extra pins system is made and connected with sensors so when the
die touches the sensor it will stop the bending depending on the angle desired. It is only necessary to put the
small pin in the hole desired in order to activate the system but there are no calculations related to ir (even the
pin is not shown in the final assembly because it consists of a different mechanism and it does not affect the
general functioning of the assembly.
As we can see and appreciate in the following figure, the pin system is just to have a representation and not
to leave it as it is.
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Figure 16. Pin system reference.
Now, for the bigger pin which holds the moving die and the holders, it is a piece of 1 inch aluminum which
due its size and even though it will hold certain amount of force, it is assumed that it can resist the force
being applied because the forces acting on it are even on all of the surface and its moving with all of the
rotary system itself.
Just to be sure, we proceed to calculate the tensile strength of the bolt using the following formula
from reference [24] and which is:
𝑇 = 𝐴 * 𝑆 * 𝐹 (15)
Where:
A is the cross-sectional area of the bolt
S is the ultimate tensile strength of the bolt material
F is the factor of safety
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Collecting the data we know that the diameter of our bolt is 1 inch which gives us an area of 0.00543 square
feet, the ultimate tensile strength of the 4130 is 120,000 psi and we is is used a safety factor of 3 which is the
recommended value for the project itself (increased to 5 to be sure).
Finally, after applying these values into the formula we obtain that:
𝑇 = 0. 00543 * 120, 000 * 5 = 3, 258 𝑝𝑜𝑢𝑛𝑑𝑠 𝑜𝑟 1, 478 𝐾𝑔 (15)
If we multiply this value times the gravity itself to get the force that it can hold about 14,500 kg and knowing
that our system will produce a force of 3,600N, it is known for a fact that the pin can hold the force and it can
be used for the proposed system.
Finally, to end the calculations, the supports holders that will transmit the systems force in order for the die to
bend the tube. These holders have a thickness of 2 cm each and we will assume that they can resist the force
being applied due their size and thickness (it will be later confirmed with the use of FEM analysis).
EXTRA INFORMATION REGARDING THE TABLE´S LEGS ATTACHMENT:
As it can be seen, the welding of the legs is shown in order to show reference of how it will be used.
There is a PTR profile along the top surface and is
welded to the table of all of the surfaces in contact in
order to have a better and stronger structure.
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6. Finite Element Simulation Analysis
For the further analysis in the program ANSYS it is necessary to say that all the materials used were plugged
in in the system in order to get an accurate representation of the calculations and to see if each piece was
going to tolerate and hold the force being applied to bend the tube. Also, the force applied was the one
needed to bend the tube and a little bit more just in case to be sure it will work.
For the mesh size, it was tried three different mesh sizes to see which one will be the most accurate to
use due limitations on the software license, the following result for the test are shown:
Mesh Size Comparison
Optimal
Default
Worst
As it can be seen, there are even big changes in the representation of the forces and how they can be
perceived meaning that we will use the Optimal mesh size which is 0.009m size. Even though, on every
analysis there will be conducted a more precise analysis and explanation for each arangement.
Finally, the analysis of each piece is presented below with its respective total deformation, von-mises,
equivalent stress and its elastic analysis for each of the pieces with its own little analysis in order to
corroborate if the piece was going to succeed in its trial.
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Shaft (2) analysis in ANSYS program with respective force
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Spring 2023
After the FEM analysis it can be concluded that the total deformation is about .1 mm which in comparison of
all of the data collected it is the highest value obtained but even though this value can be kind of high for a
mechanical element that will be exposed to this force, it is also important to notice that this deformation
comes from the motor itself that is making the shaft rotate meaning that it won't stay in its place like this
simulation reducing the stress and the deformation will be way more smaller.
The equivalent elastic strain value is about .0003 m/m and again, it does not represent any harm to the
shaft because it is presented on the changes in size of the diameters and it is a way of the object to release the
stress through those places. In general, the system will be rotating meaning that again, the stress will be
smaller than the represented above.
Finally, the equivalent stress and the maximum principal stress as it can be seen, is located on similar
areas than the elastic strain but have (just by looking at the results) low values such as the 6.8 - 7-8 E+7Pa on
average which in comparison to the material's resistance of 505E+8 which is way bigger than those values,
we can conclude that the shaft can and will resist all of the forces of the system.
Again, it is important to state that the shaft will be rotating with the force itself and the analysis was
done as if the shaft was static due software limitations increasing the values obtained and meaning that in
reality, the shaft could possibly resist more than the calculations stated above. “Note that the mesh size was
0.009 as mentioned before with a nodes count of 824 in total”.
Hex bolt (3) analysis in ANSYS program with respective force
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For the Bolt previously calculated, it is assumed that the bolt is fixed in the part where it has contact on the
motor support while the force being applied relies where the motor is applying its own weight. The weight is
divided in 4 because as it is presented, 4 bolts are being used to hold the motor itself in the support.
The total deformation as we can see is .05 mm and on the other side of the bolt which is near to
nothing meaning that there is almost no deformation in the bolt at all, this deformation may be to the
configuration itself and that the bolt can be related as a cantilever.
On the equivalent elastic strain we can also see that there is not much of a big number being a
maximum of .06 which is a disposable value for the system itself. In general, the bolt is in the color blue
which means that all of the structure is not subjected to the max stress.
Finally, the equivalent stress and the maximum principal stress is between 1.2E+6 Pa and 9.8E+5
respectively and we see not much of stress in the bolt itself making it resistant to further fatigue while
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holding the motor. The motor will be subject to some movement due to the nature of it but it can be ignored
because the stresses are not big enough for the bolt to suffer any permanent damage.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 5,300 in total”.
Hex nut (4) analysis in ANSYS program with respective force
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Spring 2023
Moving onto the Nut, there is no previous calculation as stated because the own seller recommends the best
fit for the bolt used so the inclusion of this analysis is merely to see if the seller is right according to their
standards and if it can hold the weight and pressure that it is being applied to. Starting with the total
deformation, it is the smaller deformation that we are seeing until this point being .4 micrometers (almost
close to a minimum deformation).
The deformation itself is not a danger for the nut so we assume it is a perfect fit for the bolt and for
the system itself. Same case for the Elastic strain as we don't see a major problem regarding the distribution
of the values obtained.
The von Mises analysis shows a 5.4 mpa stress while the maximum principal has a value of 4.
Compared to the 250 mPa a stainless steel yield strength (approximation) can hold, there is nothing to worry
about making it a good fit as mentioned before for the system and implying that there is no need to worry
about the nuts in the future.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 2,090 in total”.
Motor support (5) analysis in ANSYS program with respective force
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In comparison to the previous analysis as it can be seen and in general for the system, the mesh size can
look way smaller that other parts, it is important to clarify that this is due the size of the pieces making
the mesh as a matter of size being smaller. The mesh size could be smaller for some parts including the
shaft but due limitations of the software we needed to stay at the mesh size stated above.
The total deformation is about .1 micrometers and this can be to the optimal welding configuration on
the sides of the figure (previously calculated and vaidated), also the size and the material could help to
reduce this value which is almost disposable. Same case for the elastic stress which only on the sides
where the weld is is creating certain stress, but not a stress where it needs to be worried about so we
proceed to not look at the values at all of .4 micrometers.
For the values obtained in the von-Mises and the maximum principal analysis, it is known that
.28 and .18 mPa (even more smaller that the previous stresses obtained) which if we refer again to the
yield stress of the material in general, the stress generated to the system does not represent any harm or
possible threat to it making it safe to carry on.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 9403 in total”.
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Table (5.1) analysis in ANSYS program with respective force
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On the table side, it is interesting to see that in general on all of the analysis the stresses and deformations are
not affecting our system at all, which is perfect because it is being referred to as the main structure that is
holding the weight at all. The legs of the table are the ones being affected the least and in some cases there is
0 deformation so there is no problem regarding them.
Being more specific, the total deformation is 1.4E-10 which as stated makes it near to 0 and we won't
worry about it. This only says that the ptr profiles and the aluminum plates that compose the system are
strong enough to hold the total weight being applied.
The von-Mises and the Maximum principal are so low that it does not have an exponential on its
value which can be ignored. The only thing to see is how it is evenly distributed on the table itself and how
on the tube holders it is the most stress which is normal because they will hold the tube still so the tube will
not move.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 56,170 in total”.
Welded and rotative disc (7 - 7.1) analysis in ANSYS program with respective force
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The rotatory and the welded disc were analyzed together because we can say they are so close and work
together that we can take them as one piece ( with their respective constraints such as the bottom one being
welded and the top one as one free rotative piece). As it may be seen in most cases the bottom part of the
pieces is the lowest value because it is the part being welded while the edge of the top one is the one having
more stress and deformation.
Continuing with the deformation itself and the elastic strain, we see no important value which means
that there is no harm to the disc that will help to relieve the weight of the system to the motor. It is only now
necessary to observe the distribution shown such as the edges, the ones that have more deformation and the
even distribution of the weight along the top surface.
The von-Mises analysis again give us a .1 mPa next to the 13,795 Pa which in general terms means
nothing to the steel part meaning that the proposed arrangement where one welded disc and one rotatory one
holding the weight of part of the system it is a good idea and later on it only has to be checked that the
lubricant system is being applied correctly (important to note that also the edges of the discs are the ones that
have more of the stress applied).
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 302 in total”.
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Main die (8) analysis in ANSYS program with respective force
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Starting with the Main die which is the one that will stay still while the tube is being bended into it, the total
deformation does not represent any major or important harm to the die itself because a deformation of .6
micrometers is nothing compared to the big die size and the strong material as the h13 itself which on the
distribution of them its being applied to where the tube is supposed to go.
The stresses are about .5 mPa for the von-Mises and .4 mPa for the maximum principal but does not
represent again any harm and possible fatigue damage through time making the die calculations correct and
reliable. Moving to the distribution of these stresses is a similar case as the deformation, where the tube goes
are the most stresses but a peculiar peak of stress on the edge of the die is presented but this may be to the
arrangement in the ANSYS program where the force being applied is located along the side (including those
edges).
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 4698 in total”.
Moving die (8.1) analysis in ANSYS program with respective force
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There will not be more detailed information regarding this die because it is made of the exact same material
and size of the main die, the only difference is that the hole of it is smaller and this die is the one applying
pressure in the tube that is being bent. The smaller hole gives it more strength to hold and if we compare the
values obtained there is almost the same stress and deformation to the main die and even the stress
distribution making them similar and redundant to talk about.
The deformation is about 60 nanometers and the stress does not pass 1 mPa which means as stated
that this value does not represent any future harm or threat to the system.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 4698 in total”.
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Support pin (11) analysis in ANSYS program with respective force
Almost close to the end of the analysis, the pin is an important part of the system because it is the one that is
holding the moving die with both holders as we can see on the arrangement proposed. Moving to the total
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deformation, it gives a result of 6 micrometers which basically represents nothing to the piece.
Note that similar to other pieces, the pin is rotating with the system itself and not a still piece as
observed in the program that may cause a bigger number of deformation (even though it is not an important
one). The main deformation is seen on the middle of the part and it is decreasing its value as close to the edge
but it only means that the die is the one that applies the main force while bending the tube.
On the von-Mises side, it is seen that opposite to the deformation, the major stresses are located at the
edge of the pin which is curious because it is the opposite of what one would expect. Which means is that the
structure is searching for a way to release this stresses and what it does is it moves them to its edges, Also the
values obtained are not an important thread to the pin which is why we chose a 4130 (very strong material) in
order to hold all the forces in case if needed.
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 120 in total”.
Holders (12) analysis in ANSYS program with respective force
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At last for the individual analysis, the holders (both have the same design, same material and same force
configuration due the symmetrical system) are the ones that connect the shaft with the bearings and the
moving die with the pink analyzed above. As it may be seen, the bottom and top sides of the holders do not
represent any deformation because all of the force is being located where the bearing/shaft is and the pin
itself.
Similar to the previous pieces, the deformation and the stresses of 3.4 and 2.4 mPa for the von-Mises
and the Maximum principal respectively also does not represent any harm because the material strength helps
it to carry this force in order to bend the tube as expected. The movement of the bearings transmitting the
force of the shaft and the pin holding this together is what makes the colors inside the holes which again,
look like light colors which means that the stress is no problem at all (even the red color is a way small value
for the system itself).
“Note that the mesh size was 0.009 as mentioned before with a nodes count of 561 in total”.
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Mechanism analysis in ANSYS program with respective force
Based on the results of the FEM analysis, it can be concluded that the total deformation, von-Mises stress,
and maximum principal stress values are within safe limits, indicating that the whole movement system is not
at risk of failure or harm.
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Then, FEM analysis allows for the simulation of various loading and boundary conditions, which can
help identify potential failure modes and optimize the design of the system to prevent them. This ensures that
the system is robust and can withstand the expected operating conditions.
Finally, the forces and stresses we see on the whole system are very different compared to the
individual ones but, in general the main and critical parts such as the holes of the holders, the shaft edge
connected to the motor, the edge of the disc and the pin itself have bigger values and colors in general which
stills represent the same tendency as analyzed above. The only thing different is the total deformation which
as we see the side of the die represents the greater value of 10 cm which is a lot.
Is this correct? kind of because the system itself is moving and in the time proposed for the fem
analysis that's the movement it will have and it is not being taken as a solid cantilever part such as other
analysis mentioned. That's the reason for the big value and as it goes to the rotational edge the value
decreases a lot.
In conclusion, all of the results are between the acceptable values mentioned and it can be said that the
system can still hold all of the forces necessary to bend a tube which can lead to a successful mechanism.
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7. Functionality Analysis (attached on video archive)
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As our shaft moves, all of the system that holds it together moves with its respective rotation as we
can see the shaft (orange color) transmitting the movement while the dies rotate and they can stop whenever
they hit the pins depending on where it is placed (the angle needed). The central die only moves in its axis
while the other die is the one pushing the tube and bending it with the force transmitted from the motor to the
shaft and from the shaft to the rest of the system (important to see that que bending die is hold together with
the joint pin as seen).
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In order to have a better understanding of the movement, the following QR codes can provide a better visual
explanation.
Main system movement without table:
Main system movement with table:
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8. Documentation and project support.
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References
[1]
Benders,
B.
S.
(2017,
19
septiembre).
A
History
of
Metal
bending.
https://www.barnshaws.com/information/articles/a-history-of-metal-bending
[2] History of the Code of Ethics for Engineers | National Society of Professional Engineers. (s. f.).
https://www.nspe.org/resources/ethics/code-ethics/history-code-ethics-engineers
[3] Code of Ethics for Engineers. (2019). En NSPE, NSPE. NSPE. Recuperado 30 de enero de 2023, de
https://www.nspe.org/sites/default/files/resources/pdfs/Ethics/CodeofEthics/NSPECodeofEthicsforEn
gineers.pdf
[4] Wammes Machinery. (s. f.). 3-roll-bending machines Wammes Machinery. Wammes Machinery
GmbH||Your
partner
for
plant
and
machine
construction.
https://www.wammesmachinery.com/en/machines/round-and-rectangular-fittings-production/beadingrounding-flanging/3-roll-bending-machines
[5]
Corporation,
S.
M.
&
SMC.
(s.
f.).
Shanghai
Metal
Corporation.
https://es.shanghaimetal.com/index.php?ac=article
[6]
How
to
Make
the
Most
of
Your
Small
Bathroom
Space.
(s.
f.).
Bender.
https://benderplumbing.com/bender-blogs?id=1189277/how-to-make-the-most-of-your-small-bathroo
m-space
[7] Tubos. (s. f.). Stainless Steel hollow Tubes Size, SS Round hollow tubing Manufacturer.
https://www.tubos.in/stainless-steel-hollow-tube-manufacturer.html
[8] PIPE BENDING METHODS, TOLERANCES, PROCESS AND MATERIAL REQUIREMENTS. (2015,
diciembre). IHS MARKIT STANDARTS STORE. Recuperado 6 de febrero de 2023, de
https://global.ihs.com/doc_detail.cfm?document_name=PFI%20ES24&item_s_key=00085204
74
Spring 2023
[9] AS130: Tube Bending Radius - SAE International. (2008). SAE INTERNATIONAL. Recuperado 6 de
febrero de 2023, de https://www.sae.org/standards/content/as130/
[10] MACHINERY SAFETY STANDARDS. (2020). ANSI B11. Recuperado 6 de febrero de 2023, de
https://www.b11standards.org/current-standards
[11]
UNE-EN
ISO
13857:2020
Seguridad
de
las
máquinas.
Distancias
de.
.
.
(s.
f.).
https://www.une.org/encuentra-tu-norma/busca-tu-norma/norma?c=N0064556
[12]
Y14.5
(SPANISH)
Dimensiones
y
Tolerancias
-
ASME.
(s.
f.).
https://www.asme.org/codes-standards/find-codes-standards/y14-5-(spanish)-dimensiones-y-tolerancias-(1)
[13]
Stainless
Steel
Pipe
–
Hongqi
Iron
and
Steel
Co.,
Ltd.
(s.
f.).
https://hongqiiron.com/stainless-steel-pipe/?gclid=Cj0KCQiA54KfBhCKARIsAJzSrdpOWqBNyH6yf0gcI_
3ulusXo1AQvIHUQQkQL4_n89a0ufVBpiQmkGYaAnYBEALw_wcB
[14] Aceros y Metales Cuautitlán. (2022, 13 septiembre). ➢ Placa de aluminio aleaciones 1100, 3003, 6061
T-6, 7075 T-6. https://acerosymetalescuautitlan.com.mx/venta/placa-de-aluminio/
[15] AMOB Group. (2023, 13 enero). CNC Pipe Bender - CH Series Pipe Bending Machine.
https://www.amobgroup.com/en/products/cnc-pipe-bender/
[16] CNC Full Automatic Pipe Bending Machine_Zhangjiagang Yuetai Machinery Science Technology Co.,
Ltd. (s. f.).
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https://www.yuetaimachine.com/viewprut-440.html?gclid=Cj0KCQiA54KfBhCKARIsAJzSrdqEiSd9xS35ya
SMn2FYg007ek7VQcjjKO0S3LDisnbhr41b7Y-ETXkaAgluEALw_wcB
[17] Co., S. L. S. M. P. (2022, 20 julio). Welded. Shandong Shunhe Metal Product CO.,LTD.
https://www.shandongshunhemetal.com/products/stainless_steel_pipes/welded_steel_pipes/welded.ht
ml
[18] 1. MECHANICS OF MATERIALS Fourth Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T.
DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2006 The McGraw-Hill
Companies, Inc. All rights reserved. 5 Analysis and Design of Beams for Bending
[19] Shigley, J. E., & Mischke, C. R. (1989). Mechanical Engineering Design. McGraw-Hill.
[20] Budynas, R. G., & Nisbett, K. J. (2011). Shigley's mechanical engineering design. McGraw-Hill.
[21] Richard G. and J. Keith (2012). Shigley’s Mechanical Engineering Design (9th ed.). McGraw-Hill.
[22] Shigley, J. E., Mischke, C. R., & Brown Jr, T. H. (2004). Standard handbook of machine design.
McGraw-Hill Education.
[23] Beer, F. P., Johnston Jr., E. R., DeWolf, J. T., & Mazurek, D. F. (2017). Mechanics of Materials (7th
ed.). McGraw-Hill Education.
[24] Bickford, J. H. (1990). Handbook of bolts and bolted joints. CRC Press.
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Appendix
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Class problems (statistics):
3- Draw a scatterplot for the tube diameters and the force required to bend the tubes for the system proposed
and at least 6 different diameters. Then, calculate the correlation coefficient betwen the tubes diameters have
and the force required to bend the tubes. Finally, find the regression line that relates the tube diameters and
the force needed to bend the tubes.
First, we have to use the formula used above in the calculations (1) in order to obtain the force needed
to bend a tube, then, we plug in 6 different diameters in m to obtain the following forces in Newton (the
material of the tube will be an AISI304 with the same values proposed on table 13).
Figure 17. Comparison between the force needed to bend the tube against its diameter with a thickness of .0024m.
Then, we calculate the regression line that shows its linearly (if it has one) but as it may be seen, it clearly
shows certain linearly which is calculated and represented as it follows:
𝑟=
𝑆𝑥𝑦
𝑆𝑥𝑆𝑦
= 0. 99997
Finally, with the results we get that the relation is linear by 99% and when plotted, we get the following
regression line equation that represents the line as y = 3404.74x + 116.7115.
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Figure 18. Comparison between the force needed to bend the tube against its diameter with a thickness of .0024m and its
regression line.
----------------------------------------------------------------------------------------------------------------------------------------------------
4- As it may seem, four different companies bought the machine produced by Carlos INC which helps them
to bend about 400 tubes daily, the question is that some companies state that one of them is producing more
tubes while some of the others are producing less bendings. Can the provided data conclude that that
affirmation is correct? Perform a chi analysis for the four companies to check if they are differing on the
production of bending tubes. A significative value of 𝛂=0.05.
Keep in mind that the proposals are:
Ha: The number of tubes bended are the same for the 4 companies
Ho: The number of tubes bended are not the same for the 4 companies
then, the observed data is presented:
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Finally, we get the CHI value as 6.8125 and as the degrees of the table used we get that the CHI value in
which we are going to compare it, is 3.8414. Then as observed, the CHI value obtained is greater than the
needed with a significant value of 0.05 meaning that our first proposal is correct.
In conclusion, Ha is correct and we conclude that the 4 companies are producing the same tubes,
maybe the variation that they are claiming can depend on different variables that they are not taking into
consideration
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PATENT
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