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1. Chapter 1 Introduction to Mechanical Engineering Design Lecture Slides © 2020 McGraw Hill. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw Hill.

2. Chapter Outline 1–1 Design 4 1–2 Mechanical Engineering Design 5 1–3 Phases and Interactions of the Design Process 5 1–4 Design Tools and Resources 8 1–5 The Design Engineer’s Professional Responsibilities 10 1–6 Standards and Codes 12 1–7 Economics 13 1–8 Safety and Product Liability 15 1–9 Stress and Strength 16 1–10 Uncertainty 16 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 1–14 Dimensions and Tolerances 27 1–15 Units 31 1–16 Calculations and Significant Figures 32 1–17 Design Topic Interdependencies 33 1–18 Power Transmission Case Study Specifications 34

3. 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.

4. 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 .

5. The Design Process Iterative in nature. Requires initial estimation, followed by continued refinement. Fig. 1–1 Access the text alternative for slide images.

6. Design Considerations Some characteristics that influence the design 1 Functionality. 2 Strength/stress. 3 Distortion/deflection/stiffness. 4 Wear. 5 Corrosion. 6 Safety. 7 Reliability. 8 Manufacturability. 9 Utility. 10 Cost. 11 Friction. 12 Weight. 13 Life. 14 Noise. 15 Styling. 16 Shape. 17 Size. 18 Control. 19 Thermal properties. 20 Surface. 21 Lubrication. 22 Marketability. 23 Maintenance. 24 Volume. 25 Liability. 26 Remanufacturing/resource recovery.

7. 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 .

8. 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. 8

9. A Few Useful Internet Sites www.globalspec.com www.engnetglobal.com www.efunda.com www.thomasnet.com www.uspto.gov

10. 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.

11. 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 .

12. 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.

13. 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.

14. Standards and Codes 2 Some organizations that establish standards and codes of particular interest to mechanical engineers : Aluminum Association (AA ). American Bearing Manufacturers Association (ABMA ). American Gear Manufacturers Association (AGMA ). American Institute of Steel Construction (AISC ). American Iron and Steel Institute (AISI ). American National Standards Institute (ANSI ). American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE ). American Society of Mechanical Engineers (ASME ). American Society of Testing and Materials (ASTM ). American Welding Society (AWS ). ASM International. British Standards Institution (BSI ). Industrial Fasteners Institute (IFI ). Institute of Transportation Engineers (ITE ). Institution of Mechanical Engineers ( IMechE ). International Bureau of Weights and Measures (BIPM ). International Federation of Robotics (IFR ). International Standards Organization (ISO ). National Association of Power Engineers (NAPE ). National Institute for Standards and Technology (NIST ). Society of Automotive Engineers (SAE ).

15. 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.

16. Tolerances Close tolerances generally increase cost. Require additional processing steps. Require additional inspection. Require machines with lower production rates. Fig. 1–2 Access the text alternative for slide images. Source: From Ullman, David G., The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003 .

17. 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. Fig. 1–3 Access the text alternative for slide images .

18. 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 .

19. 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.

20. 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 .

21. 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.

22. Uncertainty 3 Deterministic method. Establishes a design factor, n d . Based on absolute uncertainties of a loss-of-function parameter and a maximum allowable parameter . If, for example, the parameter is load, then

23. 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 From Equation (1–2), the maximum allowable load is found to be Answer

24. 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 . 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 .

25. 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 = πd 2 ∕4, σ = P ∕ A , and from Equation (1–3), σ = S ∕ n d , then Solving for d yields Answer From Table A–17, the next higher preferred size is Thus , when n d is replaced with n in the equation developed above, the factor of safety n is Answer Thus , rounding the diameter has increased the actual design factor .

26. 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, σ nom = 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 = 32 M ∕ ( π d 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

27. 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 = k e d , where k e is a constant. Thus, Equation (1) is rewritten as Applying the tolerances to achieve the maximum the stress can reach gives 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 27

28. 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 = 2 P . 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 2 P for P and the maximum stress is The minimum strength is Equating the maximum stress to the minimum strength gives From Equation (1–3), the design factor using nominal values should be Answer

29. 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, n i , for each. For this example, the list would be Loss-of-Function Estimated Accuracy n i 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

30. 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.

31. Probability of Failure Probability of Failure , p f – 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.

32. Gaussian (Normal) Distribution 1 The probability density function (PDF) of the Gaussian distribution is expressed in terms of its mean, m x , and its standard deviation Fig. 1–4 Typical plots of the Gaussian distribution

33. Gaussian (Normal) Distribution 2 p f is obtained by integrating Eq. (1–4). The variable x is placed in dimensionless form using the transformation variate. The transformation variate is normally distributed, with a mean of zero and a standard deviation and variance of unity .

34. 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 .

35. 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 ( z α ). Fig. 1–5 Access the text alternative for slide images.

36. Table A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution (partial) Z α .00 .01 .02 .03 .04 .05 .06 .07 .08 .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 Access the text alternative for slide images .

37. Example 1–4 (1) In a shipment of 250 connecting rods, the mean tensile strength is found to be and has a standard deviation of Assuming a normal distribution, how many rods can be expected to have a strength less than S = 39.5 kpsi ? How many are expected to have a strength between 39.5 and 59.5 kpsi ? Solution Substituting in Equation (1–5) gives the transform z variable as The probability that the strength is less than 39.5 kpsi can be designated as F ( z ) = Φ( z 39.5) = Φ(−1 . 10). Using Table A–10, and referring to Figure 1–6, we find Φ( z 39.5 ) = 0 . 1357. So the number of rods having a strength less than 39.5 kpsi is , Answer because Φ( z 39.5 ) represents the proportion of the population N having a strength less than 39.5 kpsi . Figure 1–6 Access the text alternative for slide images .

38. Example 1–4 (2) ( b ) Corresponding to S = 59 . 5 kpsi , we have Referring again to Figure 1–6, we see that the probability that the strength is less than 59.5 kpsi is F ( z ) = Φ( z 59.5 ) = Φ(2 . 90). Because the z variable is positive, we need to find the value complementary to unity. Thus, from Table A–10 The probability that the strength lies between 39.5 and 59.5 kpsi is the area between the ordinates at z 39.5 and z 59.5 in Figure 1–6. This probability is found to be Therefore the number of rods expected to have strengths between 39.5 and 59.5 kpsi is Answer Fig. 1–6

39. Discrete Mean and Standard Deviation x i is the value of an event ( i = 1, 2, … k ). f i is the class frequency, or number of times the event x i occurs within the class frequency range. The discrete mean is, The discrete standard deviation is,

40. 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 : S ut ( kpsi ) 63.5 66.5 69.5 72.5 f 2 2 3 2 where the values of S ut 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 fx 2 63.5 2 127 8 064.50 66.5 2 133 8 844.50 69.5 3 208.5 14 480.75 72.5 2 145 10 513.50 ∑ 9 613.5 41 912.25

41. Example 1–5 (2) Class Midpoint x , kpsi Class Frequency f Extension fx Extension fx 2 63.5 2 127 8 064.50 66.5 2 133 8 844.50 69.5 3 208.5 14 480.75 72.5 2 145 10 513.50 ∑ 9 613.5 41 912.25 From Equation (1–6 ), Answer From Equation (1–7 ), Answer

42. Reliability 1 The reliability is related to the probability of failure by Example: If 1000 parts are manufactured, with 6 of the parts failing, the reliability is

43. 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. Example: A shaft with two bearings having reliabilities of 95% and 98% has an overall reliability of, R = R 1 R 2 = 0.95 (0.98) = 0.93 or 93%

44. 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 .

45. 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 .

46. Probability Density Functions 2 Mean values of stress and strength are “Average” design factor is Margin of safety for any value of stress σ and strength S is The overlap area has negative margin of safety Fig. 1–7( a )

47. 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 .

48. 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 , To find probability that m > 0, form the transformation variable of m and substitute m= 0.

49. Normal-Normal Case 2 Comparing Fig. 1–7 b with Table A–10, we see that To relate to the design factor, divide each term on the right side of Eq. (1–10) by m s and rearrange :

50. Normal-Normal Case 3 Define coefficients of variance for strength and stress. Rewrite Eq. ( e ), Squaring and solving for the design factor , The plus sign is for R > 0.5, and the minus sign for R ≤ 0.5.

51. 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 .

52. Example 1–6 A round cold-drawn 1018 steel rod has 0.2 percent mean yield strength with a standard deviation of 5.90 kpsi . The rod is to be subjected to a mean static axial load of with a standard deviation of 4.1 kip. Assuming the strength and load have normal distributions, what value of the design factor corresponds to a reliability of 0.999 against yielding? Determine the corresponding diameter of the rod . Solution For strength , For stress , 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 From Table A–10, for R = 0.999, z = −3.09. Then, Equation (1–12) gives Answer The diameter is found deterministically from Solving for d gives Answer

53. 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,

54. 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 .

55. 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.

56. 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 .

57. Choice of Dimensions 2 Four examples of which dimensions to specify Fig. 1–9 Access the text alternative for slide images .

58. 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 .

59. 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 ?

60. Example 1–7 (2) Solution ( a ) The mean value of w is given by Answer For equal bilateral tolerances, the tolerance of the gap is Answer Then, w = 0.005 ± 0.010 in, and Thus, both clearance and interference are possible. ( b ) If w min is to be 0.003 in, then , Thus , Answer

61. Units 1 The symbolic units equation for F = ma is, 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 .

62. 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.

63. Gravitational System of Unit Two standard gravitational systems used by engineers. U.S. customary foot-pound-second system (fps ) . 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 ) . The derived unit of mass lbf  s 2 /in has no official name .

64. Absolute System of Units International System of Units (SI ). Base units are meter, kilogram, and second. Derived unit is newton.

65. 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 ,

66. 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 6–51 3–96 5–65 6–52 3–97 5–66 6–53 3–98 5–67 *Each row corresponds to the same mechanical component repeated for a different design concept .

67. 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

68. 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

69. 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

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