machine design design

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Page 1 Machine Design Design What you’ll learn: What’s design? What’s important in a good design? What’s fatigue? How important is a fatigue failure? What’s S-N diagram? What’s safety factor? Why we need to define safety facto for our design? Motivation: Have you ever thought that the equipment you are using for any specific function could be better? If it doesn’t work as well as you expected, did you discover what the problem is? Did you start to give some ideas to make it better? Did you try some of your ideas? Definition: Design: is to create through a series of decision makings. Engineering Design: is the process of applying the various techniques and scientific principles for the purpose of defining a device, a process or a system in sufficient detail to permit its realization. The process of good decision making: 1. Keep an Open Mind. (Don’t fall in love with your first ideas) 2. Formulation of the Problem: Determining what precisely is the problem to be solved. If the problem is overwhelmingly complex, break it down until a relatively clear understanding of the several related but smaller problems is achieved. This is not a trivial process. Indeed it is key to good decision making. 3. Systems Modeling: Determining the factors that will substantially affect outcomes and assessing the interaction of these factors through a systems approach. These factors generally fall into two major categories:

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Page 1: Machine Design Design

Page 1

Machine Design

Design

What you’ll learn:

What’s design?

What’s important in a good design?

What’s fatigue?

How important is a fatigue failure?

What’s S-N diagram?

What’s safety factor?

Why we need to define safety facto for our design?

Motivation:

Have you ever thought that the equipment you are using for any specific function could be

better?

If it doesn’t work as well as you expected, did you discover what the problem is?

Did you start to give some ideas to make it better?

Did you try some of your ideas?

Definition:

Design: is to create through a series of decision makings. Engineering Design: is the process of applying the various techniques and scientific principles for the purpose of defining a device, a process or a system in sufficient detail to permit its realization. The process of good decision making:

1. Keep an Open Mind. (Don’t fall in love with your first ideas) 2. Formulation of the Problem: Determining what precisely is the problem to be solved. If

the problem is overwhelmingly complex, break it down until a relatively clear understanding of the several related but smaller problems is achieved. This is not a trivial process. Indeed it is key to good decision making.

3. Systems Modeling: Determining the factors that will substantially affect outcomes and assessing the interaction of these factors through a systems approach. These factors generally fall into two major categories:

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Variables that parameterize physical quantities. (This is where fundamental scientific knowledge plays a role.)

Factors that, though not easily quantifiable, are, in most cases, more critical than those in the first category. Examples of these second types include:

– Culture and ethnicity – Globalization – Ethics – Economics – Political landscapes – Etc.

4. Risk Assessment: Determining the probabilistic factors that influence outcomes, for example, a lack of knowledge about the factors in (2) above and the associated costs of this gap. Some risks are obvious and well documented in the engineering and standardization literature (designs for specific life expectancy, for example). Others are much more difficult to predict. Engineers should be able to take risks without an unreasonable fear of failure.

5. Team Work and Communication: Communicating well and demonstrating the ability and willingness to work in teams. Engineers must be flexible, adaptable, and resilient.

6. Problem solving: Bringing creativity to the challenge of problem solving. Engineers must possess a high degree of creativity. Problem solving is not a rigorous walk on a deterministic path.

7. Judgment: Evaluating various options and trade-offs and identifying the best possible solution. Whereas in science there are unique answers to deterministic problems in engineering there is no globally best “design.” For example, there are at least ten different designs for wine bottle openers, each with its own function (and beauty).

Design steps

Now it’s your turn. Get in groups and talk with your group mates about design steps and make a list of your steps. There are a lot of ideas for design steps. Some of them are more detailed and some of them have fewer steps. Following is a suggested list of design steps. You can compare these steps with yours and complete whatever you missed in your list.

1. Problem Identification: Get with Customer and defined and vague problem statement.

2. Background research: Information is necessary to fully define and understand the problem

3. Goal statement: Restate the goal in a more reasonable and realistic way than in the original problem statement

4. Task specification: Creation of the detailed set of task specification which bonds the problem and limit its scope.

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5. Conceptual Design: Ideas, Sketches and Solution Lists. Make a list of The largest possible number of creative solutions

6. Analysis: Possible solutions from previous step are analyzed and either accepted, rejected or modified. Use Computer Modeling, Data Base Development and etc.

7. Selection: The most promising solution is selected. 8. Prototyping: Visualizing and Improving the Design. 9. Detailed design: Complete engineering drawings made, vendors identified,

manufacturing specifications defined. 10. Prototyping and testing: The first actual construction of the working design 11. Production: Final mass production

As you realized a design process requires a large amount of knowledge and it’s impossible unless doing in team. Regardless of what you are designing you need to do all of these steps. You will experience this when you are doing your project at the second half of this course. Figure 1 demonstrates some of these steps.

One of the very important parts of this procedure is step 6. You need to know how to analyze different parts of your equipment. What are your tools to do this? In the rest of this lessens we will study some of the subjects which will help you to analyze your design. You will practice the other steps in you project at the second half of the semester.

Figure 1. Design Graphics

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Failure

Example:

1. Versailles train crash http://en.wikipedia.org/wiki/Versailles_train_crash

2. Liberty Ship http://en.wikipedia.org/wiki/Liberty_ship

Why do parts fail? This is a question that have occupied scientists and engineers for centuries. You may answer this question based on what you have learned so far like this “pars fail because their stress exceeds their strength”. You are right up to the point, but this answer arises some other questions, likes: what kind of stress will cause that? Is there any difference between different materials? Do all materials fail because of normal stress or shear stress? Dose this failure depend on the character of loading (whether static or dynamic)? The short answer to these questions is “yes. It does matter.” But, how? You need to be patient. You will find out all of these answers in your future courses like machine design. Since here there is no enough time to cover all of these subjects we will answer some of the very important question. We will try to find out what’s failure? What is fatigue? And how the loading types can affect our design?

Earlier you get to know about the yield stress. When you design an engineering structure you want your structure could resist the applied load and when the load removed it can return to its original shape. The members must not fracture, because this would lead to a catastrophic failure that would result in material and financial loss and perhaps the loss of human life. Therefore, members in most structures are designed to sustain a maximum stress that is below the yield stress on the stress-strain diagram for the particular material used to construct that member. This maximum stress is called the design stress or allowable stress. When a properly designed member is subjected to a load, the stress in the member will not exceed the design stress. Because the design stress is within the elastic range of the material, the member will return to its original dimensions after the load is removed. A bridge, for example, sustains stresses in its members while traffic passes over it. When there is no traffic, the members in the bridge return to their original size. Similarly, while a boiler is operating, the pressure vessel sustains stresses that deform it, but when the pressure is reduced to atmospheric pressure, the vessel returns to its original dimensions [1]. But the challenging question is how to define design stress and the real stress in the structure’s members? There are a lot of parameters which affect both of these stresses. As we studied earlier, stress concentration is one the most vital factors affecting stress on different parts of a structure depending on the geometry of the member. Design stress will vary by different factors. Such as, temperature, surface condition, reliability, load characteristic, etc. Between these entire factors load characteristic has a prim importance. It introduces a new fact ‘fatigue’.

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Fatigue Failure

Fatigue is “the progressive and localized structural damage that occurs when a material is subjected to cyclic loading”. The maximum stress values are less than the ultimate tensile stress limit, and may be below the yield stress limit of the material. So your design will be totally different if you have a static load or dynamic load. Most failures in machinery are due to time-varying loads rather than to static loads. These failures typically occur at stress levels significantly lower than the yield strengths of the material. Thus, not being concern about load characteristic will lead us to unsafe designs. This phenomenon was first noticed in the 1800s when railroad-car axels began failing after only limited time in service. A German engineer, August Wohler, made the first scientific investigation over a 12 year period. He tested axels in the laboratory under fully reversed loading. He introduced a curve “The S-N” or “Wohler” as shown in Figure 2. In this curve the material strength verses the number of cycles are presented. This diagram became the standard way to characterize the behavior of materials under reversed loading.

Creating estimated S-N diagram This is a graph of the magnitude of a cyclical stress (S) against the logarithmic scale of cycles to failure (N). S-N curves are derived from tests on samples of the material to be characterized where a regular sinusoidal stress is applied by a testing machine which also counts the number of cycles to failure. In this process each test generates one point on the plot. Repeating test would complete the curve. As you may notice plotting S-N curve is just based on experiment and to have the S-N curve for each single material these tests should be done. This is the area of mechanical engineering which we don’t have laws and basic definitions like Newton’s second law or stress, strain definition. Here we just rely on experiment results and it would be different for different material and application. The S-N diagram for the particular material and application has 2 regimes. Low Cycle Fatigue (LCF) with less than 103 cycles and High Cycle Fatigue (LCF) from 103 to 106 cycles and beyond. To plot an estimated S-N diagram we need to draw a log-log axis (x axis). Based on material we know Sut . Sut is the material ultimate stress. This is the amount of stress which material can resist without failing with static force. However this one will be reduced by having cyclic force. The material strength at 103 cycles called Sm. Test data indicate that the following estimate for Sm is reasonable.

Axial loading Sm=0.75 Sut The x axis is runs from N=100 to 109 cycles or beyond. The curve starts with Sut at N=100 . The Sm is plotted at N=103. Se is the material strength at N= 106 cycles. This stress is estimated by experiment (these data are published in material handbooks). For instance: Steels Se = 0.5 Sut for Sut < 200kpsi (1400MPa) Se = 100 kpsi (700 MPa) for Sut ≥ 200kpsi (1400MPa) Irons Se = 0.4 Sut for Sut < 60kpsi (400MPa) Se = 24 kpsi (160 MPa) for Sut ≥ 60kpsi (400MPa)

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Now, we need to draw a straight line between Sut , Sm and Sm, Se. Depend on material, usually S-N curve exhibits a knee. In this case the curve is continued horizontally beyond Se. If your machine is designed to work for 104 cycles, its fatigue strength would be determined by the S-N curve (Figure 2).

Se

Sm

Figure 2. S-N diagram

Fatigue point ς

Sut

103 106

Sn

N 104

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Example: Create an S-N diagram for a steel bar. How many cycles of life can expect if the alternating stress is 350 MPa. The Sut has been tested at 600 MPa. Solution: Estimate Sm and Se based on ultimate strength. Sm= 0.75 Sut = 0.75 (600) = 450 MPa Steels Se = 0.5 Sut = 0.5 (600) =300 MPa for Sut < 200kpsi (1400MPa) The approximate number of cycles of life can be found from the diagram, which is approximately 105

cycles.

Sut = 600

103 106

Sn

N

Sm = 450

Se = 300 ς = 350

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Safety factor

Because design is not an exact science, engineers incorporate an allowance in their designs. That’s why we need a safety factor. To account for the uncertainties, engineers use a design or allowable stress based on a parameter called the factor of safety or safety factor. This factor can measure the quality of design. It can be expressed in many different ways. It is typically a ratio of two quantities that have the same units, such as strength/stress, critical load/applied load, maximum cycles/applied cycles. A safety factor is always unitless. Maximum load that a structure member or a machine component will be allowed to carry under normal conditions of utilization is considerably smaller than the ultimate load. This smaller load is referred to as the allowable load and, sometimes, as the working load or design load. Thus, only a fraction of the ultimate load capacity of the member is utilized when the allowable load is applied. The remaining portion of the load caring capacity of the member is kept in reserve to assure its safe performance. Here the factor of safety (F S.) is defined the ratio of the failure stress to the allowable stress:

F.S. = failure stress / allowable stress Because the yield stress is the stress above which a material plastically deforms, the yield stress is commonly used as the failure stress. The ultimate stress may also be used. The failure stress is always greater than the allowable stress, so F.S. > 1. The value chosen for the factor of safety depends on the type of engineering structure, the relative importance of the member compared with other members in the structure, the risk to property and life, and the severity of the design uncertainties previously listed. For example, the safety factor of Commercial air craft is 1.2 - 1.5. However military air craft can have safety factor less than 1.1, but their crews wear parachute. Missiles safety factor is 1 but have no crew and aren’t expected to return anyway. These small factors of safety in aircraft are necessary to keep weight low and are justified by sophisticated analytical modeling. However, the factor of safety for ground-based structures such as dam, bridges, and buildings may be higher, perhaps 1.5 or 2. High-risk structures that pose a safety hazard to people in the event of failure, such as certain nuclear power plant component may have a factor of safety as high as 3. Factors of safety for structural members in specific engineering systems have been standardized through many years of testing and industrial evaluation. Many engineering societies and government agencies have developed codes for specific areas of engineering design. Most are only recommendations, but some have the force of law. The ASME provides recommended guidelines for safety factors to be used in particular applications such as steam boilers and pressure vessels. Building codes are legislated in most U.S.A. states and cities and usually deal with publicity accessible structures or their components, such as elevators and escalators. Safety factors are sometimes specified in these codes and may be quite high (the code for escalators in one state called for a factor of safety of 14). Clearly where human safety is involved high values of N are justified. However they come with a weight and cost penalty as parts must often be made heavier to achieve large values of N. the design

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engineer must always be aware of these codes and standards and adhere to them where applicable.

Example:

A concrete column with a diameter of 60 cm supports a portion of a highway overpass. Using the ultimate stress as the failure stress, what is the maximum compressive load that the column can carry for a factor of safety of 1.25? For the ultimate stress of concrete use 40 MPa. Solution:

Safety factor is failure stress / allowable stress: F.S. = ςfail ς𝑎𝑙𝑙𝑜𝑤 = ςu ς𝑎𝑙𝑙𝑜𝑤 = 1.25

ς𝑎𝑙𝑙𝑜𝑤 = ςu 1.25 = 40 (𝑀𝑃𝑎) 1.25 = 32 𝑀𝑃𝑎 = 32 (106 𝑁 𝑚2 ) ςal low = Fmax A The column has a circular area.

A = π * r2 Radius = r = diameter / 2 = 60 (cm) / 2 = 30 cm 𝐴 = 𝜋𝑟2= 𝜋 (30 𝑐𝑚)2 = 2826 𝑐𝑚2 = 2826 (10−2 𝑚)2 = 2826 ∗ 10−4𝑚2 Fmax = ςallow ∗ A = 32 106 𝑁 𝑚2 ∗ 2826 ∗ 10−4𝑚2 = 90432 ∗ 102𝑁

= 9.04 ∗ 106𝑁 = 9.04 𝑀𝑁

Example:

Design a flat bar with a transverse hole which is subjected to a 3 MN force in axial direction.

Given:

F = 3MN axial tension

Kt = 2.5

S.F. = 2

Cycles N=103

8 cm<d< 10cm

Geometry (w = 50 cm, d = ? cm, h = 30 cm, D=50 cm)

Financial limitations (choose the cheapest material)

Material:

1. Aluminum 1100 cold rolled: (Sut = 165 MPa), 20 $ / kg, Density (ρ = 2800 kg/m3) 2. Aluminum 43 permanent mold casting: (Sut = 159 MPa), 19 $ / kg, Density (ρ =

2700 kg/m3)

D

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Solution:

Material 1:

Sm = 0.75 Sut = 0.75 (165) = 123.75 MPa

ςmax = S. F. ∗ Kt ∗ ςnom

𝜍𝑛𝑜𝑚 =𝑆𝑚

𝐾𝑡 ∗ 𝑆. 𝐹.=

(123.75 𝑀𝑃𝑎)

2.5 ∗ 2 = 24.75 𝑀𝑃𝑎

𝜍𝑛𝑜𝑚 =𝑃

𝐴

𝐴 =𝑃

𝜍𝑛𝑜𝑚=

3 ∗ 106 𝑁

24.75 ∗ 106 𝑃𝑎= 0.12121𝑚2

A = w * h – d * h = (w-d)*h

d= 0.50-(A/0.3)= 0.50- (0.12121/0.3)= 0.096m = 9.6 cm

Volume of shaft ∶ 𝑉 = 𝐷 ∗ 𝑤 ∗ ℎ − 𝜋𝑟2h

V = (0.5)(0.3)(0.5) – (3.14)(0.0962)(0.3)/4 = 0.0728 m3

𝑚𝑎𝑠𝑠 = 𝜌 ∗ 𝑉 = 2800 ∗ 0.0728 = 203.92𝑘𝑔

𝑝𝑟𝑖𝑐𝑒 = 𝑚𝑎𝑠𝑠 ∗$

𝑘𝑔= 203.92 ∗ 20 = 4078 $

Material 2:

Sm = 0.75 Sut = 0.75 (159) = 119.25 MPa

ςmax = S. F. ∗ Kt ∗ ςnom

𝜍𝑛𝑜𝑚 =𝑆𝑚

𝐾𝑡 ∗ 𝑆. 𝐹.=

(119.25 𝑀𝑃𝑎)

2.5 ∗ 2 = 23.85 𝑀𝑃𝑎

𝜍𝑛𝑜𝑚 =𝑃

𝐴

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𝐴 =𝑃

𝜍𝑛𝑜𝑚=

3 ∗ 106 𝑁

23.85 ∗ 106 𝑃𝑎= 0.1258 𝑚2

A = w * h – d * h = (w-d)*h

d= 0.50-(A/0.3)= 0.50- (0.1258/0.3)=0.0807 m = 8.07 cm

Volume of shaft ∶ 𝑉 = 𝐷 ∗ 𝑤 ∗ ℎ − 𝜋𝑑2h/4

V = (0.5)(0.3)(0.5) – (3.14)(0.08072)(0.3)/4 = 0.0735 m3

𝑚𝑎𝑠𝑠 = 𝜌 ∗ 𝑉 = 2700 ∗ 0.0735 = 198.36𝑘𝑔

𝑝𝑟𝑖𝑐𝑒 = 𝑚𝑎𝑠𝑠 ∗$

𝑘𝑔= 198.36 ∗ 18 = 3767 $

Since both calculated diameters are between 8 cm and 10 cm, both of them are acceptable and

we can choose the cheapest one, which is the second one.

Designed bar is: Aluminum 43 permanent mold casting

Geometry (w = 50 cm, d = 8 cm, h = 30 cm, D=50 cm)

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References

1. Kirk D. Hagen, “Introduction to Engineering Analysis”, 2nd Edition, Prentice Hall 2. Robert L. Norton, “Machine Design”, 3rd edition, Prentice Hall