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Chapter 1
Introduction to Mechanical
Engineering Design
Lecture Slides
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Chapter Outline
1–1
Design 4
1–2
Mechanical Engineering
Design 5
1–3
Phases and Interactions of the
Design Process 5
1–11 Design Factor and Factor of
Safety 18
1–12 Reliability and Probability of
Failure 20
1–13 Relating Design Factor to
Reliability 24
1–4
Design Tools and Resources 8
1–5
The Design Engineer’s
Professional Responsibilities 10
1–14 Dimensions and Tolerances 27
1–6
Standards and Codes 12
1–7
Economics 13
1–16 Calculations and Significant
Figures 32
1–8
Safety and Product Liability 15
1–9
Stress and Strength 16
1–10 Uncertainty 16
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1–15 Units 31
1–17 Design Topic
Interdependencies 33
1–18 Power Transmission Case
Study Specifications 34
2
Design
To formulate a plan for the satisfaction of a specified need.
Process requires innovation, iteration, and decision-making.
Communication-intensive.
Products should be.
• Functional.
• Safe.
• Reliable.
• Competitive.
• Usable.
• Manufacturable.
• Marketable.
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3
Mechanical Engineering Design
Mechanical engineering design involves all the disciplines of
mechanical engineering.
Example.
• Journal bearing: fluid flow, heat transfer, friction, energy
transport, material selection, thermomechanical treatments,
statistical descriptions, etc.
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4
The Design Process
Iterative in nature.
Requires initial estimation,
followed by continued
refinement.
Access the text alternative for slide images.
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Fig. 1–1
5
Design Considerations
Some characteristics that influence the design
1
2
3
4
5
6
7
8
9
10
11
12
13
Functionality.
Strength/stress.
Distortion/deflection/stiffness.
Wear.
Corrosion.
Safety.
Reliability.
Manufacturability.
Utility.
Cost.
Friction.
Weight.
Life.
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15
16
17
18
19
20
21
22
23
24
25
26
Noise.
Styling.
Shape.
Size.
Control.
Thermal properties.
Surface.
Lubrication.
Marketability.
Maintenance.
Volume.
Liability.
Remanufacturing/resource recovery.
6
Computational Tools
Computer-Aided Engineering (CAE).
• Any use of the computer and software to aid in the engineering
process.
• Includes.
• Computer-Aided Design (CAD).
• Drafting, 3-D solid modeling, etc.
• Computer-Aided Manufacturing (CAM).
• CNC toolpath, rapid prototyping, etc.
• Engineering analysis and simulation.
• Finite element, fluid flow, dynamic analysis, motion, etc.
• Math solvers.
• Spreadsheet, procedural programming language, equation solver, etc.
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Acquiring Technical Information
Libraries.
• Engineering handbooks, textbooks, journals, patents, etc.
Government sources.
• Government agencies, U.S. Patent and Trademark, National
Institute for Standards and Technology, etc.
Professional Societies (conferences, publications, etc.)
• American Society of Mechanical Engineers, Society of
Manufacturing Engineers, Society of Automotive Engineers, etc.
Commercial vendors.
• Catalogs, technical literature, test data, etc.
Internet.
Access to much of the above information.
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A Few Useful Internet Sites
www.globalspec.com
www.engnetglobal.com
www.efunda.com
www.thomasnet.com
www.uspto.gov
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The Design Engineer’s Professional Responsibilities
Satisfy the needs of the customer in a competent, responsible,
ethical, and professional manner.
Some key advise for a professional engineer.
• Be competent.
• Keep current in field of practice.
• Keep good documentation.
• Ensure good and timely communication.
• Act professionally and ethically.
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Ethical Guidelines for Professional Practice
National Society of Professional Engineers (NSPE) publishes a
Code of Ethics for Engineers and an Engineers’ Creed.
www.nspe.org/ethics.
Six Fundamental Canons.
Engineers, in the fulfillment of their professional duties, shall:
• 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.
• Conduct themselves honorably, responsibly, ethically, and lawfully so
as to enhance the honor, reputation, and usefulness of the profession.
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NSPE Engineers’ Creed
As a Professional Engineer I dedicate my professional knowledge
and skill to the advancement and betterment of human welfare.
I pledge:
• To give the utmost of performance;
• To participate in none but honest enterprise;
• To live and work according to the laws of man and the highest
standards of professional conduct;
• To place service before profit, the honor and standing of the
profession before personal advantage, and the public welfare
above all other considerations.
In humility and with need for Divine Guidance, I make this pledge.
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Standards and Codes 1
Standard.
• A set of specifications for parts, materials, or processes.
• Intended to achieve uniformity, efficiency, and a specified quality.
• Limits the multitude of variations.
Code.
• A set of specifications for the analysis, design, manufacture, and
construction of something.
• To achieve a specified degree of safety, efficiency, and performance or
quality.
• Does not imply absolute safety.
Various organizations establish and publish standards and codes for
common and/or critical industries.
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Standards and Codes 2
Some organizations that establish standards and codes of particular
interest to mechanical engineers:
Aluminum Association (AA).
ASM International.
American Bearing Manufacturers Association
(ABMA).
British Standards Institution (BSI).
American Gear Manufacturers Association
(AGMA).
Institute of Transportation Engineers (ITE).
American Institute of Steel Construction (AISC).
Industrial Fasteners Institute (IFI).
Institution of Mechanical Engineers (IMechE).
American Iron and Steel Institute (AISI).
International Bureau of Weights and Measures
(BIPM).
American National Standards Institute (ANSI).
International Federation of Robotics (IFR).
American Society of Heating, Refrigerating and
Air-Conditioning Engineers (ASHRAE).
International Standards Organization (ISO).
American Society of Mechanical Engineers
(ASME).
American Society of Testing and Materials
(ASTM).
American Welding Society (AWS).
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National Association of Power Engineers
(NAPE).
National Institute for Standards and
Technology (NIST).
Society of Automotive Engineers (SAE).
14
Economics
Cost is almost always an important factor in engineering design.
Use of standard sizes is a first principle of cost reduction.
Table A–17 lists some typical preferred sizes.
Certain common components may be less expensive in stocked
sizes.
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Tolerances
Close tolerances
generally increase cost.
• Require additional
processing steps.
• Require additional
inspection.
• Require machines
with lower production
rates.
Access the text alternative for slide images.
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Fig. 1–2
Source: From Ullman, David G., The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003.
16
Breakeven Points
A cost comparison between two possible production methods.
Often there is a breakeven point on quantity of production.
EXAMPLE
Automatic screw machine.
• 25 parts/hr.
• 3 hr setup.
• $20/hr labor cost.
Hand screw machine.
• 10 parts/hr.
• Minimal setup.
• $20/hr labor cost.
Breakeven at 50 units.
Access the text alternative for slide images.
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Fig. 1–3
17
Safety and Product Liability
Strict Liability concept generally prevails in U.S.
Manufacturer is liable for damage or harm that results because of a
defect.
Negligence need not be proved.
Calls for good engineering in analysis and design, quality control,
and comprehensive testing.
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Stress and Strength
Strength.
• An inherent property of a material or of a mechanical element.
• Depends on treatment and processing.
• May or may not be uniform throughout the part.
• Examples: Ultimate strength, yield strength.
Stress.
• A state property at a specific point within a body.
• Primarily a function of load and geometry.
• Sometimes also a function of temperature and processing.
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Uncertainty 1
Common sources of uncertainty in stress or strength.
•
•
•
•
•
•
•
•
•
•
•
•
Composition of material and the effect of variation on properties.
Variations in properties from place to place within a bar of stock.
Effect of processing locally, or nearby, on properties.
Effect of nearby assemblies such as weldments and shrink fits on stress
conditions.
Effect of thermomechanical treatment on properties.
Intensity and distribution of loading.
Validity of mathematical models used to represent reality.
Intensity of stress concentrations.
Influence of time on strength and geometry.
Effect of corrosion.
Effect of wear.
Uncertainty as to the length of any list of uncertainties.
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Uncertainty 2
Stochastic method.
• Based on statistical nature of the design parameters.
• Focus on the probability of survival of the design’s function
(reliability).
• Often limited by availability of statistical data.
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Uncertainty 3
Deterministic method.
• Establishes a design factor, nd.
• Based on absolute uncertainties of a loss-of-function parameter
and a maximum allowable parameter.
loss-of-function parameter
nd 
maximum allowable parameter
(1 - 1)
• If, for example, the parameter is load, then
loss-of-function load
Maximum allowable load 
nd
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(1 - 2)
22
Example 1–1
Consider that the maximum load on a structure is known with an uncertainty of
±20 percent, and the load causing failure is known within ±15 percent. If the load
causing failure is nominally 2000 lbf, determine the design factor and the
maximum allowable load that will offset the absolute uncertainties.
Solution
To account for its uncertainty, the loss-of-function load must increase to 1 ∕ 0.85,
whereas the maximum allowable load must decrease to 1 ∕ 1.2. Thus to offset the
absolute uncertainties the design factor, from Equation (1–1), should be
Answer
1 0.85
nd 
 1.4
1 1.2
From Equation (1–2), the maximum allowable load is found to be
Answer
© McGraw Hill
Maximum allowable load 
2000
 1400 lbf
1.4
23
Design Factor Method
Often used when statistical data is not available.
Since stress may not vary linearly with load, it is more common to
express the design factor in terms of strength and stress.
loss-of-function strength
S
nd 

allowable stress
 (or  )
(1 - 3)
All loss-of-function modes must be analyzed, and the mode with
the smallest design factor governs.
Stress and strength terms must be of the same type and units.
Stress and strength must apply to the same critical location in the
part.
The factor of safety is the realized design factor of the final design,
including rounding up to standard size or available components.
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Example 1–2
A rod with a cross-sectional area of A and loaded in tension with an axial force of P =
2000 lbf undergoes a stress of σ = P∕A. Using a material strength of 24 kpsi and a design
factor of 3.0, determine the minimum diameter of a solid circular rod. Using Table A–17,
select a preferred fractional diameter and determine the rod’s factor of safety.
Solution
Since A = πd2∕4, σ = P∕A, and from Equation (1–3), σ = S ∕ nd, then

P
P
S


A  d 2 4 nd
Solving for d yields
Answer
 4 Pnd 
d 
  S 
12
 4  2000 3 

   24 000 
12
 0.564 in
From Table A–17, the next higher preferred size is 85 in  0.625 in. Thus, when nd is
replaced with n in the equation developed above, the factor of safety n is
Answer
n
 Sd 2
4P

  24 000 0.6252
4  2000
 3.68
Thus, rounding the diameter has increased the actual design factor.
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Example 1–3 (1)
A vertical round rod is to be used to support a hanging weight. A person will place the
weight on the end without dropping it. The diameter of the rod can be manufactured
within ±1 percent of its nominal dimension. The support ends can be centered within ±1.5
percent of the nominal diameter dimension. The weight is known within ±2 percent of the
nominal weight. The strength of the material is known within ±3.5 percent of the nominal
strength value. If the designer is using nominal values and the nominal stress equation,
σnom = P∕A (as in the previous example), determine what design factor should be used so
that the stress does not exceed the strength.
Solution
There are two hidden factors to consider here. The first, due to the possibility of eccentric
loading, the maximum stress is not σ = P∕A (review Chapter 3). Second, the person may
not be placing the weight onto the rod support end gradually, and the load application
would then be considered dynamic.
Consider the eccentricity first. With eccentricity, a bending moment will exist giving
an additional stress of σ = 32M ∕ (πd 3) (see Section 3–10). The bending moment is given
by M = Pe, where e is the eccentricity. Thus, the maximum stress in the rod is given by

© McGraw Hill
P 32 Pe
P
32 Pe



A d 3 d 2 4 d 3
(1)
26
Example 1–3 (2)
Since the eccentricity tolerance is expressed as a function of the diameter, we will write
the eccentricity as a percentage of d. Let e = ked, where ke is a constant. Thus, Equation
(1) is rewritten as

4 P 32 Pke d 4 P


1  8ke 
d 2
d 3
d 2
(2)
Applying the tolerances to achieve the maximum the stress can reach gives
 max
4 P 1  0.02
 4P 
1  8  0.015   1.166  2 

2 
 d 
  d 1  0.01 
(3)
 1.166 nom
Suddenly applied loading is covered in Section 4–17. If a weight is dropped from a
height, h, from the support end, the maximum load in the rod is given by Equation (4–59)
which is
 hk 
F  W  W 1  
 W
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27
Example 1–3 (3)
where F is the force in the rod, W is the weight, and k is the rod’s spring constant. Since
the person is not dropping the weight, h = 0, and with W = P, then F = 2P. This assumes
the person is not gradually placing the weight on, and there is no damping in the rod.
Thus, Equation (3) is modified by substituting 2P for P and the maximum stress is
 max  2 1.166  nom  2.332  nom
The minimum strength is
S min  1  0.035 S nom  0.965 S nom
Equating the maximum stress to the minimum strength gives
2.332  nom  0.965 S nom
From Equation (1–3), the design factor using nominal values should be
Answer
© McGraw Hill
nd 
S nom
 nom
2.332

 2.42
0.965
28
Example 1–3 (4)
Obviously, if the designer takes into account all of the uncertainties in this example and
accounts for all of the tolerances in the stress and strength in the calculations, a design factor
of one would suffice. However, in practice, the designer would probably use the nominal
geometric and strength values with the simple σ = P∕A calculation. The designer would
probably not go through the calculations given in the example and would assign a design
factor. This is where the experience factor comes in. The designer should make a list of the
loss-of-function modes and estimate a factor, ni, for each. For this example, the list would be
Loss-of-Function
Estimated Accuracy
ni
Geometry dimensions
Good tolerances
1.05
Stress calculation
Dynamic load
Bending
Not gradual loading
Slight possibility
2.0*
1.1
Strength data
Well known
1.05
*Minimum
Each term directly affects the results. Therefore, for an estimate, we evaluate the product
of each term
nd   ni  1.05  2.01.11.05  2.43
© McGraw Hill
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Reliability Method of Design
Reliability method of design – A method of design that relates the
distribution of stresses with the distribution of strengths to achieve
an acceptable success rate.
Reliability, R – The statistical measure of the probability that a
mechanical element will not fail in use.
© McGraw Hill
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Probability of Failure
Probability of Failure, pf – the number of instances of failures per
total number of possible instances.
Probability Density Function, PDF – the distribution of events
within a given range of values.
Two common PDFs.
• Gaussian (normal) distribution.
• Weibull distribution.
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Gaussian (Normal) Distribution 1
The probability density function (PDF) of the Gaussian distribution
is expressed in terms of its mean, mx, and its standard deviation ˆ x
2

1
1  x  mx  
f  x 
exp   


ˆ x 2
 2  ˆ x  
small ˆ x
(1 - 4)
large ˆ x
Fig. 1–4 Typical plots of the Gaussian distribution
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Gaussian (Normal) Distribution 2
pf is obtained by integrating Eq. (1–4).
The variable x is placed in dimensionless form using the
transformation variate.
x  mx
z
ˆ x
(1 - 5)
The transformation variate is normally distributed, with a mean of
zero and a standard deviation and variance of unity.
© McGraw Hill
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Cumulative Distribution Function for Gaussian Distribution 1
Integration of the Gaussian distribution to find the cumulative
density function F(x) is accomplished numerically.
To avoid the need for many tables for different values of mean and
standard deviation, the z transform is used.
The integral of the transform is tabulated in Table A–10.
© McGraw Hill
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Cumulative Distribution Function for Gaussian Distribution 2
A sketch of the standard normal distribution, showing the z
transform is given below.
The normal cumulative density function is labeled Φ(zα).
Fig. 1–5
  z   
z

 u2 
1
exp    du
 2
2


1  
z  0
z  0
Access the text alternative for slide images.
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Table A–10 Cumulative Distribution Function of Normal
(Gaussian) Distribution (partial)
  z   
z

 u2 
1
exp    du
 2
2


1  
z  0
z  0
Zα
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.4960
0.4920
0.4880
0.4840
0.4801
0.4761
0.4721
0.4681
0.4641
0.1
0.4602
0.4562
0.4522
0.4483
0.4443
0.4404
0.4364
0.4325
0.4286
0.4247
0.2
0.4207
0.4168
0.4129
0.4090
0.4052
0.4013
0.3974
0.3936
0.3897
0.3859
0.3
0.3821
0.3783
0.3745
0.3707
0.3669
0.3632
0.3594
0.3557
0.3520
0.3483
0.4
0.3446
0.3409
0.3372
0.3336
0.3300
0.3264
0.3238
0.3192
0.3156
0.3121
0.5
0.3085
0.3050
0.3015
0.2981
0.2946
0.2912
0.2877
0.2843
0.2810
0.2776
0.6
0.2743
0.2709
0.2676
0.2643
0.2611
0.2578
0.2546
0.2514
0.2483
0.2451
0.7
0.2420
0.2389
0.2358
0.2327
0.2296
0.2266
0.2236
0.2206
0.2177
0.2148
0.8
0.2119
0.2090
0.2061
0.2033
0.2005
0.1977
0.1949
0.1922
0.1894
0.1867
0.9
0.1841
0.1814
0.1788
0.1762
0.1736
0.1711
0.1685
0.1660
0.1635
0.1611
1.0
0.1587
0.1562
0.1539
0.1515
0.1492
0.1469
0.1446
0.1423
0.1401
0.1379
1.1
0.1357
0.1335
0.1314
0.1292
0.1271
0.1251
0.1230
0.1210
0.1190
0.1170
1.2
0.1151
0.1131
0.1112
0.1093
0.1075
0.1056
0.1038
0.1020
0.1003
0.0985
1.3
0.0968
0.0951
0.0934
0.0918
0.0901
0.0885
0.0869
0.0853
0.0838
0.0823
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Example 1–4 (1)
In a shipment of 250 connecting rods, the mean tensile strength is found to be S  45 kpsi and has a
standard deviation of ˆ S  5 kpsi.
(a) Assuming a normal distribution, how many rods can be expected to have a strength less than
S = 39.5 kpsi?
(b) How many are expected to have a strength between 39.5 and 59.5 kpsi?
Solution
(a) Substituting in Equation (1–5) gives the transform z variable as
x  m x S  S 39.5  45
z39.5 


 1.10
ˆ x
ˆ S
5
The probability that the strength is less than 39.5 kpsi can be designated as F(z) = Φ(z39.5) = Φ(−1.10).
Using Table A–10, and referring to Figure 1–6, we find Φ(z39.5) = 0.1357. So the number of rods having
a strength less than 39.5 kpsi is,
Answer
N  ( z39.5 )  250(0.1357)  33.9  34 rods
because Φ(z39.5) represents the proportion of the population N having a strength less than 39.5 kpsi.
Figure 1–6
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Example 1–4 (2)
(b) Corresponding to S = 59.5 kpsi, we have
z39.5 
59.5  45
 2.90
5
Referring again to Figure 1–6, we see that the probability that the strength is less than 59.5 kpsi is
F(z) = Φ(z59.5) = Φ(2.90). Because the z variable is positive, we need to find the value
complementary to unity. Thus, from Table A–10
 (2.90)  1   (2.90)  1  0.001.87  0.998 13
The probability that the strength lies between 39.5 and 59.5 kpsi is the area between the ordinates at
z39.5 and z59.5 in Figure 1–6. This probability is found to be
p   ( z59.5 )   ( z39.5 )   (2.90)   (1.10)
 0.998 13  0.1357  0.862 43
Therefore the number of rods expected to have strengths between 39.5 and 59.5 kpsi is
Answer
Np  250(0.862)  215.5  216 rods
Fig. 1–6
© McGraw Hill
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Discrete Mean and Standard Deviation
xi is the value of an event (i = 1, 2, … k).
fi is the class frequency, or number of times the event xi occurs
within the class frequency range.
The discrete mean is,
1 k
x   f i xi
N i 1
(1 - 6)
The discrete standard deviation is,
k
sx 
© McGraw Hill

fi xi 2  N x 2
i 1
N 1
(1 - 7)
39
Example 1–5 (1)
Five tons of 2-in round rods of 1030 hot-rolled steel have been received for workpiece stock.
Nine standard-geometry tensile test specimens have been machined from random locations
in various rods. In the test report, the ultimate tensile strength was given in kpsi. The data in
the ranges 62 to 65, 65 to 68, 68 to 71, and 71 to 74 kpsi is given in histographic form as
follows:
Sut (kpsi)
f
63.5
66.5
69.5
72.5
2
2
3
2
where the values of Sut are the midpoints of each range. Find the mean and standard
deviation of the data.
Solution
Table 1–1 provides a tabulation of the calculations for the solution.
Table 1–1
Class Midpoint x, kpsi
Class Frequency f
Extension fx
Extension fx2
63.5
66.5
69.5
72.5
2
2
3
2
9
127
133
208.5
145
613.5
8 064.50
8 844.50
14 480.75
10 513.50
41 912.25
∑
© McGraw Hill
40
Example 1–5 (2)
Class Midpoint x, kpsi
Class Frequency f
Extension fx
Extension fx2
63.5
66.5
69.5
72.5
2
2
3
2
9
127
133
208.5
145
613.5
8 064.50
8 844.50
14 480.75
10 513.50
41 912.25
∑
From Equation (1–6),
Answer
1 k
1
x   f i xi  (613.5)  68.16667  68.2 kpsi
N i 1
9
From Equation (1–7),
k
Answer
sx 

fi xi 2  N x 2
i 1
N 1
41 912.25  9(68.16667 2 )

9 1
 3.39 kpsi
© McGraw Hill
41
Reliability 1
The reliability is related to the probability of failure by
R  1 pf
(1 - 8)
Example: If 1000 parts are manufactured, with 6 of the parts
failing, the reliability is
6
R  1
 0.994
1000
© McGraw Hill
or 99.4%
42
Reliability 2
Series System – a system that is deemed to have failed if any
component within the system fails.
The overall reliability of a series system is the product of the
reliabilities of the individual components.
n
R   Ri
(1 - 9)
i 1
Example: A shaft with two bearings having reliabilities of 95% and
98% has an overall reliability of,
R = R1 R2 = 0.95 (0.98) = 0.93
© McGraw Hill
or 93%
43
Relating Design Factor to Reliability 1
Reliability is the statistical probability that machine systems and
components will perform their intended function satisfactorily
without failure.
Deterministic relations between stress, strength, and design factor
are often used due to simplicity and difficulty in acquiring
statistical data.
Stress and strength are actually statistical in nature.
© McGraw Hill
44
Probability Density Functions 1
Stress and strength are statistical in nature.
Plots of probability density functions shows distributions.
Overlap is called interference of σ and S, and indicates parts
expected to fail.
Fig. 1–7 (a)
Access the text alternative for slide images.
© McGraw Hill
45
Probability Density Functions 2
Mean values of stress and strength are   m and S  m S
“Average” design factor is
nd 
mS
m
(a )
Margin of safety for any value of stress σ and strength S is
m  S 
The overlap area has negative margin of safety
(b)
Fig. 1–7(a)
© McGraw Hill
46
Margin of Safety
Distribution of margin of safety is dependent on distributions of
stress and strength.
Reliability R is area under the margin of safety curve for m > 0.
Interference is the area 1–R where parts are expected to fail.
Fig. 1–7 (b)
Access the text alternative for slide images.
© McGraw Hill
47
Normal-Normal Case 1
Common for stress and strength to have normal distributions.
Margin of safety is m = S – σ, and will be normally distributed.
Reliability is probability p that m > 0. That is,
R  p ( S   )  p ( S    0)  p (m  0)
(c )
To find probability that m > 0, form the transformation variable of
m and substitute m=0.
m S  m
m  mm 0  mm
mm
z



ˆ m
ˆ m
ˆ m ˆ S2  ˆ 2 1 2

© McGraw Hill

(1 - 10)
48
Normal-Normal Case 2
Comparing Fig. 1–7b with Table A–10, we see that
R  1  ( z )
z0
 ( z )
z0
(d )
To relate to the design factor, nd  m S m , divide each term on the
right side of Eq. (1–10) by m and rearrange:
mS
1
m
nd  1
z

© McGraw Hill
12
 ˆ
ˆ  
 m  m2 
 
 
nd  1
2
S
2
2

12
 m S2 ˆ S2 ˆ 2 
 m2 m2  m2 
 
  S
12
 ˆ m
ˆ  
 m m  m2 
 
 
nd  1

12
 2 ˆ S2 ˆ 2 
 nd m 2  m 2 
S
 

2
S
2
2
S
2
S
2
(e )
49
Normal-Normal Case 3
Define coefficients of variance for strength and stress.
CS  ˆ S m S and C  ˆ  m
Rewrite Eq. (e),
z
nd  1
n C  C
2
d
2
S
2
(1 - 11)
Squaring and solving for the design factor,
nd 
1  1  (1  z 2CS2 )(1  z 2C2 )
1 z C
2
2
S
(1 - 12)
The plus sign is for R > 0.5, and the minus sign for R ≤ 0.5.
© McGraw Hill
50
Relating Design Factor to Reliability 2
In summary, Eq. (1–12) relates the design factor to the reliability
goal (through the transform variate) and the coefficients of
variation of strength and stress.
nd 
1  1  (1  z 2CS2 )(1  z 2C2 )
1 z C
2
2
S
(1 - 12)
CS  ˆ S m S and C  ˆ  m
© McGraw Hill
51
Example 1–6
A round cold-drawn 1018 steel rod has 0.2 percent mean yield strength S y  78.4 kpsi with a standard
deviation of 5.90 kpsi. The rod is to be subjected to a mean static axial load of P  50 kip with a standard
deviation of 4.1 kip. Assuming the strength and load have normal distributions, what value of the design factor
nd corresponds to a reliability of 0.999 against yielding? Determine the corresponding diameter of the rod.
Solution
For strength, CS  ˆ S m S  5.90 78.4  0.0753. For stress,
P 4P
  2
A d
Since the tolerance on the diameter will be an order of magnitude less than that of the load or strength, the
diameter will be treated deterministically. Thus, statistically, the stress is linearly proportional to the load, and
C  CP  ˆ P m P  4.1 50  0.082. From Table A–10, for R = 0.999, z = −3.09. Then, Equation (1–12)
gives
Answer
1  1  1  (3.09) 2 (0.0753) 2  1  ( 3.09) 2 (0.082) 2 
nd 
 1.416
1  (3.09) 2 (0.0753) 2
The diameter is found deterministically from
4P S
 2 y
d
nd
Solving for d gives
4 Pnd
4(50)(1.416)
d

 1.072 in
Answer
Sy
 (78.4)
© McGraw Hill
52
Dimensions and Tolerances 1
Nominal size – The size we use in speaking of an element.
• Is not required to match the actual dimension.
Limits – The stated maximum and minimum dimensions.
Tolerance – The difference between the two limits.
Bilateral tolerance – The variation in both directions from the basic
dimension, for example, 1.005 ± 0.002 in.
Unilateral tolerance – The basic dimension is taken as one of the
limits, and variation is permitted in only one direction, for example,
1.005
© McGraw Hill
0.004
0.000
in
53
Dimensions and Tolerances 2
Clearance – Refers to the difference in sizes of two mating cylindrical
parts such as a bolt and a hole.
• Assumes the internal member is smaller than the external member.
• Diametral clearance – difference in the two diameters.
• Radial clearance – difference in the two radii.
Interference – The opposite of clearance, when the internal member is
larger than the external member.
Allowance – The minimum stated clearance or the maximum stated
interference or mating parts.
Fit – The amount of clearance or interference between mating parts.
GD&T – Geometric Dimensioning and Tolerancing, a comprehensive
system of symbols, rules, and definitions for defining the theoretically
perfect geometry, along with the allowable variation.
© McGraw Hill
54
Choice of Tolerances
The designer is responsible for specifying tolerances for every
dimension.
Consideration is given to functionality, fit, assembly,
manufacturing process ability, quality control, and cost.
Excessive precision is a poor design choice, in that it limits
manufacturing options and drives up the cost.
Less expensive manufacturing options should be selected, even
though the part may be less than perfect, so long as the needs are
satisfactorily met.
© McGraw Hill
55
Choice of Dimensions 1
Dimensioning a part is the designer’s responsibility.
Include just enough dimensions.
Avoid extraneous information that can lead to confusion or multiple
interpretations.
Example of over-specified dimensions. With +/– 1 tolerances, two
dimensions are incompatible.
Fig. 1–8
Access the text alternative for slide images.
© McGraw Hill
56
Choice of Dimensions 2
Four examples of which dimensions to specify
Fig. 1–9
Access the text alternative for slide images.
© McGraw Hill
57
Tolerance Stack-up
The cumulative effect of individual tolerances must be allowed to
accumulate somewhere. This is known as tolerance stack-up.
Chain dimensioning allows large stack-up of many small tolerances
in series.
Baseline dimensioning minimizes large tolerance stack-up.
Access the text alternative for slide images.
© McGraw Hill
58
Example 1–7 (1)
A shouldered screw contains three hollow right circular cylindrical parts on the
screw before a nut is tightened against the shoulder. To sustain the function, the
gap w must equal or exceed 0.003 in. The parts in the assembly depicted in
Figure 1–10 have dimensions and tolerances as follows:
a = 1.750 ± 0.003 in
b = 0.750 ± 0.001 in
c = 0.120 ± 0.005 in
d = 0.875 ± 0.001 in
Fig. 1–10
All parts except the part with the dimension d are supplied by vendors. The part
containing the dimension d is made in-house.
(a) Estimate the mean and tolerance on the gap w.
(b) What basic value of d will assure that w ≥ 0.003 in?
© McGraw Hill
59
Example 1–7 (2)
Solution
(a) The mean value of w is given by
Answer w  a  b  c  d  1.750  0.750  0.120  0.875  0.005 in
For equal bilateral tolerances, the tolerance of the gap is
Answer
tw   t  0.003  0.001  0.005  0.001  0.010 in
all
Then, w = 0.005 ± 0.010 in, and
wmax  w  tw  0.005  0.010  0.015 in
wmin  w  tw  0.005  0.010  0.005 in
Thus, both clearance and interference are possible.
(b) If wmin is to be 0.003 in, then, w  wmin  t w  0.003  0.010  0.013 in. Thus,
Answer
© McGraw Hill
d  a  b  c  w  1.750  0.750  0.120  0.013  0.867 in
60
Units 1
The symbolic units equation for F = ma is,
F  MLT 2
(1 - 13)
• F – Force.
• M – Mass.
• L – Length.
• T – Time.
Any three of these units can be selected as base units.
The fourth unit is then a derived unit.
© McGraw Hill
61
Units 2
Gravitational System of Units.
• When force, length, and time are chosen as base units, and mass
is the derived unit.
• English-based systems are usually gravitational systems.
Absolute System of Units.
• When mass, length, and time are chosen as base units, and force
is the derived unit.
• Metric systems are usually absolute systems.
© McGraw Hill
62
Gravitational System of Unit
Two standard gravitational systems used by engineers.
• U.S. customary foot-pound-second system (fps).
FT 2 (pound-force)(second) 2
M

 lbf  s 2 /ft  slug
L
foot
(1 - 14)
• The unit of force is the pound, or more properly the pound-force (lbf).
• The derived unit of mass is the slug.
• Inch-pound-second system (ips).
FT 2 (pound-force)(second) 2
M

 lbf  s 2 /in
L
inch
(1 - 15)
• The derived unit of mass lbf  s2/in has no official name.
© McGraw Hill
63
Absolute System of Units
International System of Units (SI).
• Base units are meter, kilogram, and second.
• Derived unit is newton.
ML (kilogram)(meter)
2
F 2 

kg

m/s
N
2
T
(second)
© McGraw Hill
(1 - 16)
64
Units of Weight
The weight of an object is the force exerted on it by gravity.
Designating weight as W and acceleration due to gravity as g,
W  mg
© McGraw Hill
(1 - 17)
65
Linked End-Of-Chapter Problems
Table 1–2 Problem Numbers for Linked End-of-Chapter Problems*
3–1
4–50
4–81
3–41
5–78
5–79
3–79
4–23
4–29
4–35
5–50
6–37
7–12
11–16
3–80
4–24
4–30
4–36
5–51
6–38
7–13
11–17
3–81
4–25
4–31
4–37
5–52
6–39
7–14
11–18
3–82
4–26
4–32
4–38
5–53
6–40
7–15
11–19
3–83
4–27
4–33
4–39
5–54
6–41
7–16
7–24
7–25
7–39
11–29
11–30
13–44
14–37
3–84
4–28
4–34
4–40
5–55
6–42
7–17
7–26
7–27
7–40
11–31
11–32
13–45
14–38
3–85
5–56
6–43
7–18
11–47
13–48
3–87
5–57
6–44
7–19
11–48
13–48
3–88
5–58
6–45
7–20
11–20
13–46
14–39
3–90
5–59
6–46
7–21
11–21
13–47
14–40
3–91
4–41
4–78
5–60
6–47
3–92
5–61
6–48
3–93
5–62
6–49
3–94
5–63
6–50
3–95
4–43
4–80
5–64
5–67
3–96
5–65
6–52
3–97
5–66
6–53
3–98
5–67
6–51
*Each row corresponds to the same mechanical component repeated for a different design concept.
© McGraw Hill
66
Power Transmission Case Study Specifications 1
Design Requirements
Power to be delivered: 20 hp
Input speed: 1750 rev/min
Output speed: 85 rev/min
Targeted for uniformly loaded applications, such as conveyor belts, blowers,
and generators
Output shaft and input shaft in-line
Base mounted with 4 bolts
Continuous operation
6-year life, with 8 hours/day, 5 days/wk
Low maintenance
Competitive cost
Nominal operating conditions of industrialized locations
Input and output shafts standard size for typical couplings
© McGraw Hill
67
Power Transmission Case Study Specifications 2
Design Specifications
Power to be delivered: 20 hp
Power efficiency: >95%
Steady state input speed: 1750 rev/min
Maximum input speed: 2400 rev/min
Steady-state output speed: 82 to 88 rev/min
Usually low shock levels, occasional moderate shock
Input and output shafts extend 4 in outside gearbox
Input and output shaft diameter tolerance: ±0.001 in
Input and output shafts in-line: concentricity ±0.005 in, alignment ±0.001 rad
Maximum allowable loads on input shaft: axial, 50 lbf; transverse, 100 lbf
Maximum allowable loads on output shaft: axial, 50 lbf; transverse, 500 lbf
Maximum gearbox size: 14-in × 14-in base, 22-in height
Base mounted with 4 bolts
© McGraw Hill
68
Power Transmission Case Study Specifications 3
Design Specifications
Mounting orientation only with base on bottom
100% duty cycle
Maintenance schedule: lubrication check every 2000 hours; change of
lubrication every 8000 hours of operation; gears and bearing life >12 000
hours; infinite shaft life; gears, bearings, and shafts replaceable
Access to check, drain, and refill lubrication without disassembly or opening
of gasketed joints
Manufacturing cost per unit: <$300
Production: 10 000 units per year
Operating temperature range: −10 degrees to 120 degrees F
Sealed against water and dust from typical weather
Noise: <85 dB from 1 meter
© McGraw Hill
69
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