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Solutions for Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012

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PHYSICAL PROPERTIES OF MATERIALS

Review Questions 4.1

Define density as a material property. Answer. Density is the weight per unit volume of a material.

4.2

What is the difference in melting characteristics between a pure metal element and an alloy metal? Answer. A pure metal element melts at one temperature (the melting point), while an alloy begins melting at a certain temperature called the solidus and finally completes the transformation to the molten state at a higher temperature called the liquidus. Between the solidus and liquidus, the metal is a mixture of solid and liquid.

4.3

Describe the melting characteristics of a noncrystalline material such as glass. Answer. In the heating of a noncrystalline material such as glass, the material begins to soften as temperature increases, finally converting to a liquid at a temperature defined for these materials as the melting point.

4.4

Define specific heat as a material property. Answer. Specific heat is defined as the quantity of heat required to raise the temperature of a unit mass of the material by one degree.

4.5

What is thermal conductivity as a material property? Answer. Thermal conductivity is the capacity of a material to transfer heat energy through itself by thermal movement only (no mass transfer).

4.6

Define thermal diffusivity. Answer. Thermal diffusivity is the thermal conductivity divided by the volumetric specific heat.

4.7

What are the important variables that affect mass diffusion? Answer. According to Fick's first law, mass diffusion depends on the diffusion coefficient of the material, which increases rapidly with temperature (so temperature could be listed as an important variable), concentration gradient, contact area, and time.

4.8

Define resistivity as a material property. Answer. Resistivity is the material's capacity to resist the flow of electric current.

4.9

Why are metals better conductors of electricity than ceramics and polymers? Answer. Metals are better conductors because of metallic bonding, which permits electrons to move easily within the metal. Ceramics and polymers have covalent and ionic bonding, in which the electrons are tightly bound to particular molecules.

4.10 What is dielectric strength as a material property? Answer. Dielectric strength is defined as the electrical potential required to break down the insulator per unit thickness. 4.11 What is an electrolyte? Answer. An electrolyte is an ionized solution capable of conducting electric current by movement of the ions. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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Solutions for Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012

Problems Answers to problems labeled (A) are listed in an Appendix at the back of the book. 4.1

(A) (USCS units) Determine the increase in length of a steel bar whose length = 100 in, if the bar is heated from room temperature of 70F to 500F. Use Table 4.1 for reference. Solution: From Table 4.1, coefficient of thermal expansion  = 6.7(10-6)/F for steel. Increase in length = 6.7(10-6)(100)(500 - 70) = 0.288 in

4.2

(SI units) Determine the length and width of a rectangular nickel plate whose room temperature (20C) dimensions are 750 mm by 400 mm by 5 mm, if the plate is heated to 250C. Use Table 4.1 for reference. Solution: From Table 4.1, coefficient of thermal expansion  = 13.3(10-6)/C for nickel. Eq. (4.1) for length: L2 – L1 = αL1(T2 – T1) L2 = 750 + 13.3(10-6)(750)(250 – 20) = 750 + 2.294 = 752.294 mm Eq. (4.1) for width: W2 – W1 = αW1(T2 – T1) W2 = 400 + 13.3(10-6)(400)(250 – 20) = 400 + 1.224 = 752.294 mm

4.3

(A) (SI units) A steel shaft has a starting diameter = 15.00 mm. It is to be inserted into a hole in an expansion fit assembly operation. To be readily inserted, the shaft must be reduced in diameter by cooling. Determine the temperature to which the shaft must be reduced from room temperature (20C) in order to reduce its diameter to 14.98 mm. Use Table 4.1 for reference. Solution: From Table 4.1, coefficient of thermal expansion  = 13.3(10-6)/C for nickel. Eq. (4.1) for the diameter: D2 - D1 = D1 (T2 - T1). 14.98 - 15.00 = 12(10-6)(15.00)(T2 - 20) -0.02 = 180(10-6)(T2 - 20) -0.02 = 0.00018(T2 - 20) = 0.00018T2 - 0.0036 -.02 + 0.0036 = 0.00018T2 -0.0164 = 0.00018T2 T2 = -91.1C

4.4

(SI units) A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Expansion joints will be provided to compensate for the change in length in the girders as the temperature varies. Each expansion joint can compensate for a maximum of 20 mm of change in length. From historical records it is estimated that the minimum and maximum temperatures in the region will be -35°C and 40°C, respectively. Determine (a) the minimum number of expansion joints required and (b) the length that each bridge section should be fabricated. Use Table 4.1 for reference. Solution: (a) From Table 4.1, the coefficient of thermal expansion α=12(10-6)/C for steel. If the length of the bridge is 500 m at 20°C, first determine what its length would be at -35°C using Eq. (4.1): L2 – 500 = 12(10-6)(500)(-35 - 20) = -0.33 m L2 = 499.67 m at -35°C Now determine the entire expansion over the range of temperatures between -35 °C and +40°C L2 – L1 = 12(10-6) (499.67)(40 – (-35)) L2 – L1 = 0.4497 m = 449.7 mm Each expansion joint will provide 20 mm of expansion. The number of expansion joints must be 449.7/20 = 22.485 which is rounded up to 23. Therefore, a minimum of 23 joints are needed for coverage of the total bridge length. (b) If there are 23 joints separating adjacent sections of the bridge, then there must be 24 bridge sections to be fabricated. Each bridge section will be 500/24 = 20.833 m long at 20°C.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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Solutions for Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012

4.5

(USCS units) A zinc die casting has a critical dimension of 7.500 in when it solidifies in the mold at its melting temperature because 7.500 in is the corresponding mold dimension. What is the value of this part dimension when it cools to room temperature (70°F)? Use Table 4.1 for reference. Solution: From Table 4.1, the melting temperature of zinc = 787°F and its coefficient of thermal expansion = 22.2(10-6)/°F. L2 - L1 = L1 (T2 - T2) L2 = 7.500 + 22.2(10-6)(7.500)(70 - 787) = 7.500 – 0.119 = 7.381 in

4.6

(SI units) Aluminum has a density of 2.70 g/cm3 at room temperature (20C). Determine its density at 650C, using data in Table 4.1 for reference. Solution: Assume a 1 cm3 cube, 1 cm on each side. From Table 4.1,  = 24(10-6) mm/mm/C L2 - L1 = L1 (T2 - T2). L2 = 1.0 + 24(10-6)(1.0)(650 - 20) = 1.01512 cm (L2 )3 = (1.01512)3 = 1.04605 cm3 Assume weight remains the same; thus  at 650C = 2.70/1.04605 = 2.581 g/cm3

4.7

(SI units) Determine the amount of heat required to increase the temperature of an aluminum block that is 10 cm by 10 cm by 10 cm from room temperature (21C) to 300C. Use Table 4.2 for reference. Solution. Heat = (0.21 cal/g-C)(103 cm3)(2.70 g/cm3)(300C - 21C) = 158,193 cal Conversion: 1.0 cal = 4.184J, so heat = 662,196 J

4.8

(SI units) What is the resistance R of a length of copper wire whose length = 10 m and whose diameter = 0.30 mm? Use Table 4.3 for reference. Solution: R = rL/A, A = (0.30)2/4 = 0.0707 mm2 = 0.0707(10-6) m2 From Table 4.3,resistivity r = 1.7 x 10-8 -m2/m R = (1.7 x 10-8 -m2/m)(10 m)/( 0.0707(10-6) m2) = 240.5(10-2)  = 2.405 

4.9

(A) (USCS units) A 16 gage copper wire (0.0508-in diameter) connects a solenoid to a control circuit that is 50 ft away. (a) What is the resistance of the wire? Use Table 4.3 as a reference. (b) If a current was passed through the wire, it would heat up. How does this affect the resistance? Solution: (a) From Table 4.3, the resistivity of copper = 0.67(10-6) -in L = 50 ft = 600 in Area A = (0.0508)2/4 = 0.00203 in2 R = r (L/A) = 0.67(10-6)(600/0.00203) = 0.198  (b) If a current is passed through the wire causing the wire to heat up, the resistivity of the wire would change. Since nickel is a metal, the resistivity would increase, causing the resistance to increase. This, in turn, would cause slightly more heat to be generated.

4.10 Aluminum wiring was used in many homes in the 1960s due to the high cost of copper at the time. Aluminum wire that was 12 gauge (a measure of cross-sectional area) was rated at 15 A of current. If copper wire of the same gauge were used to replace the aluminum wire, what current should the wire be capable of carrying if all factors except resistivity are considered equal? Assume that the resistance of the wire is the primary factor that determines the current it can carry and the crosssectional area and length are the same for the aluminum and copper wires. Solution: The area and length are constant between the types of wires. The overall change in resistance is due to the change in resistivity of the materials. From Table 4.3: For Aluminum r = 2.8 x 10-8 -m For Copper r = 1.7 x 10-8 -m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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Solutions for Fundamentals of Modern Manufacturing, 5e (published by Wiley)  MPGroover 2012

The resistance will reduce by 1.7x10-8/2.8x10-8 = 0.607 Since I = E/R and Rcu=0.61(Ral), then Icu = 1/0.607Ial = 15/0.607= 24.7 A Comment: Note that the code value is actually 20 A due to several factors including heat dissipation and rounding down to the nearest 5 amp value.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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