ME_381_examples_10

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Oct 30, 2023

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ME381 Spring 2015 Homework #8 Due – 4/13/15 Complete all of the following problems from the textbook, preferably typed. Chapter 18 Review Questions 1, 5, 10, 16 Problems 2 Chapter 19 Review Questions 1, 5, 8, 10 Problems 6 Chapter 20 Review Questions 2, 3, 7, 10 Chapter 21 Review Questions 3, 4, 5, 15, 22 Problems 2 Chapter 18 – Review Questions 18.1 Why are abrasive processes technologically and commercially important? Answer. Important reasons include (1) applications on all types of materials, (2) very fine finishes, and (3) close tolerances. 18.5 What is wheel grade? Answer. Wheel grade refers to the wheel's ability to retain abrasive grains during cutting. It indicates the bond strength of the bonding material used to shape the wheel. A soft grade indicates that the grains are released easily from the bonding material. A hard wheel is one which retains the abrasive grains. 18.10 What is truing, in reference to grinding wheels? Answer. Truing is similar to dressing, but it also restores the ideal cylindrical shape to the wheel. It uses a diamond pointed tool fed slowly and precisely across the wheel while it rotates. 18.16 Name some of the abrasive operations available to achieve very good surface finishes. Answer. High finish abrasive processes include honing, lapping, superfinishing, buffing, and polishing. Chapter 18 – Problems 18.2 The following conditions and settings are used in a certain surface grinding operation: wheel diameter = 6.0 in, infeed = 0.003 in, wheel speed = 4750 ft/min, workspeed = 50 ft/min, and crossfeed = 0.20 in. The number of active grits per square inch of wheel surface = 500. Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work. Solution : (a) l c = ( Dd ) 0.5 = (6.0 x 0.003) 0.5 = (0.018) 0.5 = 0.1342 in (b) R MR = v w wd = (50 x 12)(0.20)(0.003) = 0.36 in 3 /min (c) n c = vwC = (4750 x 12)(0.2)(500) = 5,700,000 chips/min Chapter 19 – Review Questions

ME381 Spring 2015 Homework #8 Due – 4/13/15 19.1 Why are the nontraditional material removal processes important? Answer. Reasons for importance of nontraditional material removal processes are (1) the need to shape new metal alloys and non metals that are difficult to machine by conventional processes; (2) the requirement of unusual and complex workpart geometries; and (3) the need to avoid surface damage which is often associated with conventional machining. 19.5 What is the difference between water jet cutting, abrasive water jet cutting, and abrasive jet cutting? Answer. WJC cuts with a narrow, high velocity water stream; AWJC adds abrasive grits to the water stream; and AJM cuts with abrasive particles that have been added to a high velocity air stream. 19.8 How does increasing discharge current affect metal removal rate and surface finish in electric discharge machining? Answer. As discharge current increases, metal removal rate increases and surface finish is degraded. 19.10 Name the four principal steps in chemical machining. Answer. The four steps are (1) cleaning, (2) masking, (3) etching, and (4) demasking. Chapter 19 – Problems 19.6 The frontal working area of the electrode in an ECM operation is 2000 mm 2 . The applied current = 1800 amps and the voltage = 12 volts. The material being cut is nickel (valence = 2). (a) If the process is 90% efficient, determine the rate of metal removal in mm 3 /min. (b) If the resistivity of the electrolyte = 140 ohm-mm, determine the working gap. Solution : (a) From Table 19.1, C = 3.42 x 10 -2 mm 3 /A-s From Eq. (19.6) R MR = f r A = ( CI / A ) A = CI = (3.42 x 10 -2 mm 3 /A-s)(1800 A) = 6156 x 10 -2 mm 3 /s = 61.56 mm 3 /s = 3693.6 mm 3 /min At 90% efficiency R MR = 0.9(3693.6 mm 3 /min) = 3324.2 mm 3 /min (b) Given resistivity r = 140 ohm-mm, I = EA / gr in Eq. (19.2). Rearranging, g = EA / Ir g = (12 V)(2000 mm 2 )/(1800 A)(140 ohm-mm) = 0.095 mm Chapter 20 – Review Questions 20.2 Identify the important reasons why metals are annealed. Answer. Reasons for annealing include (1) to reduce brittleness and improve toughness, (2) to alter microstructure so that desirable mechanical properties can be obtained, (3) to soften metals for improved machinability or formability, (4) to recrystallize cold worked metals, and (5) to relieve residual stresses from prior processes. 20.3 What is the most important heat treatment for hardening steels? Answer. The most important heat treatment for steels is martensite formation by heating steel into the austenite region and quenching.

ME381 Spring 2015 Homework #8 Due – 4/13/15 20.7 Define hardenability. Answer . Hardenability is the relative capacity of a steel to be hardened by transformation to martensite. 20.10 Define precipitation hardening. Answer. Precipitation hardening is a heat treatment in which very fine particles (precipitates) are formed so that dislocation movement is blocked and the metal is thus strengthened and hardened. Chapter 21 – Review Questions 21.1 What are some of the important reasons why manufactured parts must be cleaned? A nswer . The reasons include (1) to prepare the surface for subsequent industrial processing, (2) to improve hygiene conditions, (3) to remove contaminants which might chemically react with the surface; and (4) to enhance product appearance and performance. 21.2 Mechanical surface treatments are often performed for reasons other than or in addition to cleaning. What are the reasons? Answer. Reasons for mechanical surface treatments include deburring, improving smoothness, adding luster, and enhancing surface properties. 21.3 What are the basic types of contaminants that must be cleaned from metallic surfaces in manufacturing? Answer. Basic contaminant types mentioned in the text are (1) oil and grease, (2) solid particles, such as metal chips, abrasive grits, shop dirt, and dust, (3) buffing and polishing compounds, and (4) oxide films, rust, and scale. 21.4 Name some of the important chemical cleaning methods. Answer. The chemical cleaning methods can be categorized as follows (1) alkaline cleaning, (2) emulsion cleaning, (3) solvent cleaning, (4) acid cleaning and pickling, and (5) ultrasonic cleaning. 21.5 In addition to surface cleaning, what is the main function performed by shot peening? Answer. Shot peening is primarily used to improve the fatigue strength of metals by introducing cold working the metallic surface. 21.15 How does anodizing differ from other conversion coatings? Answer. Anodizing uses electrochemical processing methods to convert the metallic surface. The best example is aluminum anodizing. 21.22 Describe the principal methods by which organic coatings are applied to a surface. Answer. The main methods include brushing and rolling, spraying, immersion (dip coating), and flow coating. Chapter 21 – Problems

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ME381 Spring 2015 Homework #8 Due – 4/13/15 21.2 A sheet metal steel part with surface area = 100 cm 2 is to be zinc plated. What average plating thickness will result if 15 amps are applied for 12 minutes in a chloride electrolyte solution? Solution : From Table 21.1, C = 4.75 x 10 -2 mm 3 /A-s, cathode efficiency E = 95%. Volume V = ECIt = 0.95(4.75 x 10 -2 mm 3 /A-s)(15 A)(12 min)(60 s/min) = 487.35 mm 3 Area A = 100 cm 2 = 10,000 mm 2 Plating thickness d = 487.35 mm 3 /10,000 mm 2 = 0.049 mm