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Fundamentals ofManufacturing Philip D Rufe, CMfgE Editor Second Edition Solutions Manual Society of Manufacturing Engineers www.sme.org Copyright © 2002 Society of Manufacturing Engineers All rights reserved, including those of translation This book, or parts thereof, may not be reproduced by any means, including photocopying, recording or microfilming, or by any information storage and retrieval system, without permission in writing of the copyright owners No liability is assumed by the publisher with respect to use of information contained herein While every precaution has been taken in the preparation of this book, the publisher assumes no responsibility for errors or omissions Publication of any data in this book does not constitute a recommendation or endorsement of any patent, proprietary right, or product that may be involved Additional copies may be obtained by contacting: Society of Manufacturing Engineers Customer Service One SME Drive, P.O Box 930 Dearborn, Michigan 48121 1-800-733-4763 www.sme.org SME staff who participated in producing this book: Philip Mitchell, Senior Editor Rosemary Csizmadia, Production Supervisor Kathye Quirk, Graphic Designer/Cover Design Frances Kania, Administrative Coordinator Printed in the United States of America PREFACE The purpose of this manual is to show how to solve the review questions found in the book Fundamentals of Manufacturing, 2nd edition The Fundamentals of Manufacturing Solutions Manual is also intended to help instructors teach individuals studying for the Fundamentals of Manufacturing Technologist (CMfgT) certification exam Detailed answers are provided for the quantitative review questions For the qualitative review questions, the manual directs readers to the location in the book where the answer can be found Fundamentals of Manufacturing Solutions Manual Solution 1.1 3x + = x − + x x − x − x = −6 − −2 x = −8 x=4 Solution 1.2 x− = x+ 2 1 1 12 x − = x + 2 6 2x − = 9x + 2x − 9x = + −7 x = 14 x = −2 Solution 1.3 x = 2x + y x y = x + 3 y x = xy + y x − xy = y x(1 − y ) = y x= 3y 1− 2y Solution 1.4 − = x x −1 2x 3 = x( x − 1) − x x −1 2x (2 x − 2)(3) − (2 x )(2) = ( x − 1)(1) 6x − − 4x = x −1 x − x − x = −1 + x=5 Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Solution 1.5 l = w − ft P = l + l + w + w = 46 ft 2(2w − ft) + 2w = 46 ft 4w − ft + 2w = 46 ft 6w = 54 ft w = ft Solution 1.6 x = 4th test grade 0.83 + 0.76 + 0.79 + x = 0.80 0.83 + 0.76 + 0.79 + x = 0.80 4 0.83 + 0.76 + 0.79 + x = 3.20 x = 3.20 − 0.83 − 0.76 − 0.79 x = 0.82 = 82% Solution 1.7 y + y = −5 y2 + y + = −b ± b − 4ac 2a a =1 y= b=5 c=5 −5 ± 52 − 4(1)(5) 2(1) x = −1.382, −3.618 x= Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Solution 1.8 ( g + 2)( g − 3) = g − 3g + g − = g2 − g − = −b ± b − 4ac 2a a =1 g= b = −1 c = −7 −(−1) ± (−1) − 4(1)(−7) 2(1) g = 3.193, −2.193 g= Solution 1.9 6r = − 19r 6r + 19r − = −b ± b − 4ac 2a a=6 r= b = 19 c=7 −19 ± 192 − 4(6)(−7) 2(6) r = 0.333, −3.5 r= Solution 1.10 A3 = 40x x x x 40 ft A2 = (30 + 2x)x 30 ft A1 = (30 + 2x)x x A4 = 40x Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Achips = 296 ft 2 [(30 + x) x ] + [40 x ] = 296 60 x + x + 80 x = 296 x + 140 x − 296 = −b ± b − 4ac 2a a=4 x= b = 140 c = −296 −140 ± 1402 − 4(4)(−296) 2(4) x = ft (only valid solution) x= Solution 1.11 l l – in × in squares Volumebox = length × width × height = 256 in.3 (l − 4)(l − − 4)(2) = 256 (l − 4)(l − 8)(2) = 256 (l − 12l + 32 ) = 256 2l − 24l + 64 = 256 2l − 24l − 192 = −b ± b − 4ac 2a a=2 l= b = −24 c = −192 −(−24) ± (−24) − 4(2)(−192) ≈ 17.49 in 2(2) w = 17.49 in − in = 13.49 in l= Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Solution 1.12 2x + 3y = 2x + y = 2x + 3y = −1(2 x + y = 5) 2x + 3y = −2 x − y = −5 y=0 x + 3(0) = x = 2.5 Solution 1.13 3x − y = −4 x + y = 4(3 x − y = 5) 3(−4 x + y = 5) 12 x − y = 20 −12 x + 15 y = 15 y = 35 y=5 x − 2(5) = x = 15 x=5 Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Solution 1.14 x − 2y + z = −2x + 3y + z = x + 3y + 2z = Convert the equations into an augmented matrix −2 −2 Arrange the rows so the leading entries are in decreasing order 2 2 1 −2 −2 Replace the third row with the sum of itself and times the second row 1 −2 −2 + 2(1) + 2(−2) + 2(1) 1 + 2(5) 1 1 −2 −1 11 Replace the second row with the sum of –1 times itself and the first row −1(1) + −1(−2) + −1(1) + −1(5) + −1 11 1 −3 −1 11 Replace the third row with the sum of times itself and the second row 1 0 1 0 −3 5(−1) + 5(3) + 1 5(11) − 3 −3 16 52 Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Convert the matrix back into equation form x + 3y + 2z = 5y + z = − 16z = 52 z= 52 13 = 16 13 = −3 12 13 5y = − − 4 −25 y= =− 5 5y + 5 13 x + 3 − + = 4 4 15 26 x= + − =− 4 4 Solution 1.15 25 x = 125 log 25 x = log125 x log 25 = log125 x= log125 log 25 x = 1.5 Solution 1.16 log x = 2log2 x = 23 x =8 Copyright © 2002 Society of Manufacturing Engineers Page of 65 Fundamentals of Manufacturing Solutions Manual Solution 33.1 From page 259, Toyota’s production system also can be called lean production Solution 33.2 From page 261, kaizen is the Japanese term for ongoing improvement involving everyone —managers and workers Solution 33.3 From page 261, one-piece flow is being used when the introduction of one unit is balanced by the completion of another unit of finished product Solution 33.4 From page 260, Takt time is the available work time per shift divided by the customer demand rate per shift Solution 33.5 From page 263, a pull system does not allow parts to be produced until authorization is received from the subsequent operation Solution 33.6 From pages 261 and 263, kanban is the Japanese term for the signal used to trigger the production of components in JIT manufacturing Solution 34.1 From page 266, special-purpose machines can provide faster production rates and high quality for some parts Solution 34.2 From page 266, a good fixture must confine a part through six degrees of freedom Solution 34.3 From page 267, synchronous assembly is performed when all the workpieces move at the same time Solution 34.4 From page 268, the product-process (cellular) layout produces one part family in a given cell Copyright © 2002 Society of Manufacturing Engineers Page 53 of 65 Fundamentals of Manufacturing Solutions Manual Solution 34.5 From page 270, preventive maintenance advocates replacement of machine components at specified time intervals Solution 34.6 N t = 15 sec At = 20 sec Pr = ? At × Pr 100 100 × N t 100 × 15 sec Pr = = = 75% 20 sec At Nt = Solution 34.7 N t = 10 sec Pa = 10% St = ? St = N t (100 + Pa ) 10 sec (100 + 10) = = 11 sec 100 100 Solution 35.1 From page 274, dependent demands are derived from the demand for other items Solution 35.2 From page 274, the principle measure related to balancing the cost of carrying inventory with the service level required is known as inventory turns Solution 35.3 A = 100,000 units S = $50 c = $10 i = 20% EOQ = ? EOQ = AS 2(100,000)(50) = = 2, 236 ic 0.20(10) Copyright © 2002 Society of Manufacturing Engineers Page 54 of 65 Fundamentals of Manufacturing Solutions Manual Solution 35.4 From page 275, ABC inventory analysis categorizes items based on their annual usage measured in dollars Solution 35.5 From page 276, to achieve zero inventory (not near-zero), the maximum acceptable defect rate from suppliers is zero Solution 35.6 From page 276, money from customer rebates flows from left to right in the supply chain Solution 36.1 From page 281, justification for purchasing and installing automation includes to improve product quality, improve productivity, improve work environment quality, or improve development time Solution 36.2 From page 281, CIM is the integration of design, engineering, manufacturing, and business functions Solution 36.3 From page 281, MEI stands for manufacturing enterprise integration Solution 36.4 From Figure 36-1, the customer is represented by level one of the MEI wheel Solution 37.1 From page 285, bandwidth is the term used to describe the number of signals that can be carried simultaneously on the same conductor Solution 37.2 From page 286, the client server provides real-time access to database and CAD files Solution 37.3 From page 286, fiberoptic cabling is least sensitive to electromagnetic radiation Copyright © 2002 Society of Manufacturing Engineers Page 55 of 65 Fundamentals of Manufacturing Solutions Manual Solution 37.4 From page 289, bridges allow two similar but different networks to communicate with each other Solution 37.5 From Figure 37-2, the star topology uses a central hub that each computer is connected to Solution 37.6 From page 291, hypertext transfer protocol (HTTP) is the protocol used to transfer information from a web server to a web browser Solution 38.1 From page 381, M03 starts the spindle clockwise Solution 38.2 From page 294, automatic programming of tools (APT) is popular for performing four- or five-axes CNC operations Solution 38.3 From page 293, the spindle of a three-axis or two-axis CNC machine is generally designated as the Z-axis Solution 38.4 From page 298, open-loop control contains no feedback Solution 38.5 From page 297, resolvers are less sensitive to vibration and temperature changes Solution 38.6 Using the G-codes and M-function from 380 and 381, respectively, the milling program will generate the letter X The program assumes that Z0 is the surface of the part Solution 39.1 From page 301, PLCs were designed to replace relays Solution 39.2 From page 301, PLCs without data communication result in islands of automation Copyright © 2002 Society of Manufacturing Engineers Page 56 of 65 Fundamentals of Manufacturing Solutions Manual Solution 39.3 From page 303, PLCs are sized by the number of I/O (inputs and outputs) Solution 39.4 From page 303, relay ladder logic has historically been the programming language for PLCs Solution 39.5 From Figure Q39-1, if all the inputs are off (low), output D will be off since either of the three branches will turn on output D For output D to be on, input A or B must be on (high) and C must be off (low) Solution 40.1 From page 305, the hydraulic power system is typically used for heavy loads Solution 40.2 From page 306, the robotic joints of motion are called degrees of freedom Solution 40.3 From page 306, adaptive control requires outside sensing to modify the robot program Solution 40.4 From page 306, the tool coordinate system has its center on the tool flange Solution 40.5 From Figure 40-4, the shape of the work envelope created by a polar-arm robot is spherical Solution 41.1 From page 309, triangulation is used to determine the exact location of an AGV Solution 41.2 From page 309, an AS/RS system produces a smaller footprint than that of conventional storage Solution 41.3 From page 310, the density of a bar code is defined by the width of the narrowest bar Copyright © 2002 Society of Manufacturing Engineers Page 57 of 65 Fundamentals of Manufacturing Solutions Manual Solution 41.4 From page 311, radio frequency identification uses a transponder attached to a pallet or part Solution 41.5 From pages 311 and 312, a machine vision system forms an image of the part Solution 42.1 From page 315, QC is reactive in responding to defects and QA is proactive in its approach to defect prevention and reliability Solution 42.2 From page 315, TQM is driven by customer needs Solution 42.3 From page 316, final inspection is an appraisal cost Solution 42.4 From page 317, the primary function of ISO standards is to facilitate international trade Solution 42.5 From page 317, if a company is ISO 9000 certified that does not guarantee that all the parts they produce are good Solution 42.6 From page 317, U.S Congress created the Malcolm Baldrige Award Solution 42.7 From page 317, customer focus and satisfaction are the Baldrige Award criteria emphasized the most Solution 43.1 From page 320, excessive tool wear produces assignable variability Solution 43.2 From page 321, two measures of dispersion are standard deviation and range Copyright © 2002 Society of Manufacturing Engineers Page 58 of 65 Fundamentals of Manufacturing Solutions Manual Solution 43.3 x = 1.875 in σ = 0.0005 in n = 1,000 Parts < 1.8745 in = ? Parts < 1.8745 in = x − σ ±σ = 68.26% 68.26% = 34.13% The percentage of parts smaller than x = 50 % σ= x − σ = 50% − 34.13% = 15.87% 15.87% (1,000 parts) = 158.7 ≈ 159 parts Copyright © 2002 Society of Manufacturing Engineers Page 59 of 65 Fundamentals of Manufacturing Solutions Manual Solution 43.4 Subgroup 10 11 12 13 x2 22.5 22.5 20.5 22.0 19.5 23.5 22.0 20.5 22.5 23.0 19.5 21.0 20.5 x1 22.0 20.5 20.0 21.0 22.5 23.0 19.0 21.5 21.5 21.5 20.0 19.0 19.5 x= Σxi 278.125 = = 21.39 13 N R= ΣRi 29.5 = = 2.27 13 N x3 22.5 22.5 23.0 22.0 22.5 21.0 22.0 19.0 20.0 22.0 21.0 21.0 21.0 x4 24.0 23.0 22.0 23.0 22.0 22.0 20.5 19.5 22.0 23.0 20.0 21.0 20.5 SUM x 22.750 22.125 21.375 22.000 21.625 22.375 20.875 20.125 21.500 22.375 20.125 20.500 20.375 278.125 R 2.0 2.5 3.0 2.0 3.0 2.5 3.0 2.5 2.5 1.5 1.5 2.0 1.5 29.5 From Table 43-4 with a subgroup size of 4: A2 = 0.729 D4 = 2.282 D3 = UCLX = x + A2 R = 21.39 + 0.729(2.27) = 23.05 LCLX = x − A2 R = 21.39 − 0.729(2.27) = 19.74 UCLR = D4 R = 2.282(2.27) = 5.18 LCLR = D3 R = 0(2.27) = Copyright © 2002 Society of Manufacturing Engineers Page 60 of 65 Fundamentals of Manufacturing Solutions Manual Solution 43.5 USL = 23.0 LSL = 19.0 USL − LSL 6σˆ R σˆ = d2 Cp = From Table 43-3 for a subgroup size of 4: d = 2.059 σˆ = R 2.27 = = 1.10 d 2.059 Cp = USL − LSL 23.0 − 19.0 = = 0.61 < 1.33 6σˆ 6(1.10) From page 328, C p < 1.33 means the process is incapable Solution 43.6 nearest specification − x 3σˆ 23.0 − 21.39 C pk = = 0.488 3(1.10) The answer is slightly different than the answer given on page 388 due to rounding C pk = Solution 44.1 From page 331, repeatability is a test of precision Solution 44.2 From page 332, calibration is the process of comparing a measuring device against a higher-order standard of greater accuracy Solution 44.3 From page 332, rules and micrometers are absolute measurement instruments Solution 44.4 From Figure Q44-1, the diameter being measured is larger than 0.075 and less than 0.100 The diameter is, therefore, 0.075 plus the thimble value or 0.006, which equals 0.081 in Copyright © 2002 Society of Manufacturing Engineers Page 61 of 65 Fundamentals of Manufacturing Solutions Manual Solution 44.5 From page 336, CMM stands for coordinate measuring machine Solution 44.6 From page 337, technique error occurs when an operator uses the correct measurement instrument but the wrong measuring procedure Solution 44.7 When designing a go/no-go gage for a shaft, the go side must be able to fit over the largest shaft that meets specifications (2.150) Using a gage tolerance of 10% and wear allowance of 5%: Total tolerance = upper limit – lower limit = 0.000 – 0.002 = 0.002 gage tolerance = 0.1 × 0.002 = 0.0002 wear allowance = 0.05 × 0.002 = 0.0001 The go side has a gage tolerance of 0.0002 applied unilaterally on the negative side Since the go side is subject to the most wear, its diameter is decreased by 0.0001 The no-go side should not be able to fit over any shaft within specifications The no-go side has a gage tolerance of 0.0002 applied unilaterally on the positive side Since the nogo side is not designed to fit over the shaft, the wear allowance is not used Final gage dimensions: Go: 2.1499+−.0000 0002 No-Go: 2.1480+−.0002 0000 Solution 44.8 Using Table 44-1, the gage tolerance for a Class Z gage is 0.00020 The go side must be inserted into the smallest hole (3.000) The go side has a gage tolerance of 0.00020 applied unilaterally on the positive side The problem does not include a wear allowance The no-go side must not be able to be inserted into any hole within specifications The no-go diameter is 3.004 with a gage tolerance of 0.00020 applied unilaterally on the negative side Final gage dimensions Go: 3.0000+−.0002 0000 No-Go: 3.0040+−.0000 0002 Copyright © 2002 Society of Manufacturing Engineers Page 62 of 65 Fundamentals of Manufacturing Solutions Manual Solution 45.1 From page 345, a person retains 20% of what is heard Solution 45.2 From page 350, the management concept in which many of the decisions affecting the operation of the plant are made at the lower levels is called decentralization Solution 45.3 From page 351 and Figure 45-6, a Gantt chart graphically depicts the progress of various tasks with shaded horizontal bars Solution 45.4 From page 352, Pareto analysis is a problem-solving tool that identifies the vital few and the trivial many Solution 45.5 From page 354, participatory management views workers as important assets of the organization Solution 45.6 From page 354, management by objectives is an example of the goal-setting theory Solution 46.1 From page 358, an employee who participates in an unlawful union-organized strike may be discharged and not be entitled to reinstatement when the strike ends Solution 46.2 From page 360 and the choices given, employee "right-to-know" laws focus on hazardous materials Solution 46.3 From Table 46-1, the minimum standing height for a walk-in freezer would be the stature of the 95th percentile male at 74.4 in (1,890 mm) Solution 46.4 From Table 46-3, the maximum sound level that employees can be exposed to during an 8-hour period without hearing protection is 90 dBA Copyright © 2002 Society of Manufacturing Engineers Page 63 of 65 Fundamentals of Manufacturing Solutions Manual Solution 46.5 From page 364, a truck seat that resonates at Hz could cause resonance of the internal organs of the passengers Solution 46.6 From page 364-365, repeatedly using a screwdriver throughout the workday could cause a cumulative trauma disorder such as carpal tunnel syndrome Solution 47.1 F = 50,000 n = 20 i = 6% P=? P = F ( P/F , i, n) ( P/F , i, n) = 1 = = 0.3118 n (1 + i ) (1 + 0.06) 20 P = 50,000(0.3118) = $15,590 Solution 47.2 P = 15,000 − 4,000 = 11,000 i= 12% year 1% = year 12 months month 12 months n = years = 36 months year A=? A = P ( A /P, i , n) ( A /P, i , n) = i (1 + i ) n 0.01(1 + 0.01)36 = = 0.0332 (1 + i ) n − (1 + 0.01)36 − A = 11,000(0.0332) = $365 Copyright © 2002 Society of Manufacturing Engineers Page 64 of 65 Fundamentals of Manufacturing Solutions Manual Solution 47.3 Option Option P = 46,000 P = 46,000 n = years n = years i = 9% A = 4,000 i = 9% EUAC1 = P ( A/P, i, n) ( A /P, i , n) = i (1 + i ) n 0.09(1 + 0.09)5 = = 0.2571 (1 + i ) n − (1 + 0.09)5 − EUAC1 = 46,000(0.2571) = $11,827 EUAC2 = P ( A/P, i, n) + 4,000 ( A /P, i , n) = i (1 + i ) n 0.09(1 + 0.09)7 = 0.1987 = (1 + i ) n − (1 + 0.09)7 − EUAC2 = 46,000(0.1987) + 4,000 = $13,140 The annualized cost of the second option ($4,000 service contract) is more expensive It is not a sound economic decision Solution 47.4 From page 370, storing work-in-process parts is a variable cost since space and carrying costs change with the amount of work-in-process parts Solution 47.5 From page 372, inspection does not add value to a product It does not add function or esteem value to a product Copyright © 2002 Society of Manufacturing Engineers Page 65 of 65 An Excellent General Knowledge Resource For Engineers and Non-Engineers Alike! This book covers the full spectrum of rudimentary topics in the fields of manufacturing engineering and manufacturing technology Because it contains the essentials, Fundamentals of Manufacturing, Second Edition is a “must-have” for: • Veteran manufacturing engineers who need to upgrade their general knowledge • Non-engineers who need to talk intelligently about manufacturing engineering and technology on their job NEW ! It is “official” information Fundamentals of Manufacturing is based on the “body of knowledge” which the Manufacturing Engineering Certification Institute has deemed mandatory for all manufacturing engineers and technologists to know Fifteen manufacturing experts, which includes educators, practitioners in the field, subject matter specialists, have checked the content for relevancy, accuracy and clarity, guaranteeing focused self-study and solid answers to questions regarding the fundamentals It’s arguably the most well-grounded and comprehensive textbook of its kind uide” udy Gering t S d ine ende omm uring Eng(MECI) c e R “ ct te al Offici e Manufaon Institu i h t t a f o fic Certi Features • • • • • • Thorough review of manufacturing fundamentals with samples and practice problems Detailed table of contents and index Referencing feature provides quick access to figures, tables, equations, problems, and solutions Equations, newly reformatted, are arranged logically according the sequence they’re presented Includes a number key to practice problems Up-to-date with current theoretical models, notably lean manufacturing Published by the Society of Manufacturing Engineers, 2001, Edited by Philip Rufe, CMfgE pp 400 (est.), 237 figures, 47 tables, 256 equations, ISBN: 0-87263-52-44 $75 [SME Member: $63} Request Order Code: PI-2942-4475 Benefits • • • • • Increased knowledge of manufacturing engineering Specific information is easy to find “User-friendly” presentation and layout makes for good retention and enjoyable reading You learn about real-life manufacturing situations through example problems in each chapter SME’s Fundamental Manufacturing Processes® videotapes naturally tie-in (see back) Parts, 45 Chapters Include Mathematics • Physics • Materials (Metals, Plastics, Composites, Ceramics) • Product Design (Engineering Drawing, GD&T, Computer-Aided Design, Tools) • Manufacturing Processes (Cutting Tool Technology, Machining, Forming, Casting, etc.) • Production Systems (Traditional Planning and Control, Lean Production, Process Engineering, Inventory Management) • Automated Systems and Control • Quality • Manufacturing Management (Labor, Safety, and Human Factors, Engineering Economics) • Appendixes Call Now to Charge by phone (800) 733-4763 am to pm, ET’ Monday through Friday Order Online:www.sme.org/bookstore ➨ Other General Reference Materials Manufacturing Processes & Materials Widely used by manufacturing engineering educators across the country, this book masterfully covers the basic processes and machinery used in the job shop, toolroom, or small manufacturing facility At the same time, it describes advanced equipment and processes used in larger production environments Questions and problems at the end of each chapter can be used as self-tests or assignments An Instructor’s Guide is available to tailor a more structured learning experience Additional resources from SME, including the Fundamental Manufacturing Processes videotape series (selected by the Wisconsin Manufacturing Curriculum Consortium as “recommended resources” supporting the competency-based objectives of its TECH SPAN curriculum) can also be used to supplement the book’s learning objectives With 31 chapters, 45 tables, 586 illustrations, 141 equations and an extensive index, Manufacturing Processes & Materials is one of the most comprehensive texts available on this subject Published by Society of Manufacturing Engineers, 2000 • By George F Schrader, Ahmad K Elshennawy • ISBN: 0-87263-517-1 • 850 pages hardcover Order Code: PI-2861-4475 • Price: $111/Member: $84 Bring the Text Alive with Fundamental Manufacturing Processes® and Manufacturing Insights® Videotapes These 25-30 minute programs are fast-paced, informationpacked & scripted by experts in the field, offering exceptional close-up views of manufacturing systems and processes in action Each Fundamental Examples of Manufacturing Processes® video Chapter Tie-Ins provides an overview of a particular technology Manufacturing Insights® videos take you directly to where innovative technologies are being employed by leading manufacturers SME Member Number _ ❏ Nonmember ❏ Mr ❏ Ms ❏ Mrs Name/Title _ Company/Division Address: Home ❏ Work ❏ City/State/Zip Phone ( ) _ Fax ( ) _ For faster service, please write complete order code number for each item Qty Item Title and Code Price ➞ ➞ Fundamental Manufacturing Processes® videos $255( $229 SME Members) Please print or type: Topic Order Code Milling & Machining Center Basics PIVT561-4475 Turning & Lathe Basics PIVT560-4475 Basics of Grinding PIVT564-4475 Cutting Tool Materials PIVT620-4475 Sheet Metal Stamping Presses PIVT632-4475 Sheet Metal Stamping Dies & Processes PIVT631-4475 Sheet Metal Shearing & Bending PIVT626-4475 Punch Presses PIVT625-4475 Forging PIVT-675-4475 Casting PIVT-673-4475 Die Casting PIVT674-4475 Powder Metallurgy PIVT660-4475 Plastic Injection Molding PIVT667-4475 Plastic Blow Molding PIVT666-4475 Welding PIVT681-4475 Heat Treating PI-VT678-4475 Chapter 21 21 21 18 21 24 20 17 23 12 23 12 23 12 23 12 22 11 25 25 24 10 13 13 26 13 12 Sales Tax: Businesses and residents in the following states, please add applicable sales tax: NY (4% after shipping), VA (4.5%), MA & OH (5%), CT, MI & PA (6%), IL & TX (6.25%), MN (6.5%), CA (7.25%) In Canada add 7% GST Subtotal Shipping: U.S., add $7.00 first book $2.00 for each additional Canadian shipping $9.00/book, $3.00 for each additional Shipping Sales Tax TOTAL Payment: ❏ Check/money order ❏ VISA ❏ MasterCard ❏ Purchase Order enclosed Acct No Exp Date Manufacturing Insights® videos $255( $229 SME Members) Composites in Manufacturing PI-VT248-4475 Implementing Just-In-Time PI-VT284-4475 Rapid Prototyping PI-VT419-4475 Ergonomics In Manufacturing PI-VT507-4475 Successful GD&T Implementation PI-VT515-4475 Cost of Poor Quality PI-VT570-4475 Introduction to Lean Manufacturing PI-VT682-4475 Visual Controls PI-VT691-4475 Mapping Your Value Stream PI-VT692-4475 14 35 17 47 17 42 33 33 33 Signature _ Mail orders to: Society of Manufacturing Engineers Attn: Customer Service Center P.O Box 6028 • Dearborn, MI 48121 Fax to SME Customer Service (313) 240-8252 or, order via regular phone: (800) 733-4763 or, online www.sme.org All prices subject to change without notice Terms: Net 30 days FOB Dearborn, Michigan Please remit in US funds drawn on US banks School purchase orders must include a line that states: Materials are being purchased for resale All books are guaranteed and may be returned within 15 days for a refund of the purchase price All videotapes are 1/2” VHS format and are not returnable unless defective Convenient Order Form on Back: ... height = 25 6 in.3 (l − 4)(l − − 4) (2) = 25 6 (l − 4)(l − 8) (2) = 25 6 (l − 12l + 32 ) = 25 6 2l − 24 l + 64 = 25 6 2l − 24 l − 1 92 = −b ± b − 4ac 2a a =2 l= b = ? ?24 c = −1 92 −(? ?24 ) ± (? ?24 ) − 4 (2) (−1 92) ≈... 2bc cos A a − b2 − c2 A = cos −1 −2bc 42 − 32 − 22 A = cos −1 ? ?2( 3) (2) = 104.5° b = a + c − 2ac cos B b2 − a − c2 B = cos −1 −2ac 32 − 42 − 22 B = cos −1... = P2 = A1 A2 F1 F2 = A1 A2 F2 = A2 F1 0 .25 0 in .2 (1,000 lb) = = 62. 5 lb in .2 A1 Solution 10.4 m s d1 = 500 mm = 0.5 m v1 = d = 400 mm = 0.4 m v2 = ? A1v1 = A2 v2 πd 12 πd (v1 ) = (v2 ) 4 d 12 (v1
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