Download Solutions Manual for Manufacturing Engineering Technology 6th Edition by Serope Kalpakjian and Steven Schmid

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Download Solutions Manual for Manufacturing Engineering Technology 6th Edition by Serope Kalpakjian and Steven Schmid

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Link full download :https://findtestbanks.com/download /solutions-manual-for-manufacturing-engineeringtechnology-6th-edition-by-serope-kalpakjian-and-stevenschmid/ Language: English ISBN-… Full description 

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Steven Schmid

Chapter 3

Physical Properties of Materials QUALITATIVE PROBLEMS

3.11 Describe the signiftcance of structures and machine components made of t materials with different coefficients of thermal expansion.

The structural fit of the machine components will depend on the thermal expan coefficient. For instance, if two materials with different thermal expansion coefficient assembled together by some means and then heated, the structure will develop inte stresses due to uneven expansion. If these stresses are high enough, the structure warp, bend, or buckle in order to balance or relieve the stresses; it will possibly some internal (residual) stresses as well. If prevented from warping, the structure develop high internal stresses which can lead to cracks. This is not always detrime  Read Free Foron 30 Days Signmay up to vote this title coefficien shrink fits are designed recognizing that materials have different  thermal expansion, and some machine elements such as thermocouples Not useful and temper  Useful  Cancel anytime. probes based on a mismatch of thermal expansion coefficients. coefficients. Special offer for students: Onlyare $4.99/month.

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Which of the properties described in this chapter

important for (a)

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Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.

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(c) Ruler: Should have low thermal expansion to maintain the measurements measurements accura and a low density to make it easy to carry.

(d) Paper clip: Should be corrosion resistant, with a stiffness that holds papers togeth without requiring requiring excessive excessive force.

(e) Music wire: Music wire, as used for guitars, is preloaded to a very high tensio order to achieve desired resonance. As such, it should have a very high strength the proper combination of stiffness and density to achieve the proper acoustics. (f) Beverage can: Should have a high thermal conductivity, low density, and good corrosion resistance.

3.13 Note in Table 3.1 that the properties of the alloys of metals have a wide ran compared with the properties of the pure metals. Explain why.

Alloying elements tend to disturb the crystal lattice of the base metal, and they do distorting the lattice by occupying lattice sites (substitutional atoms), spaces betwee tice sites (interstitials), or forming a second phase (an intermetallic compound of th elements). elements). Lattice distortion will reduce properties that depend on a repeating lattice, as thermal conductivity and melting points. Properties such as density and specific generally depend on the properties of the alloying elements, and range around the for the alloy base metal. Also, ‘alloys’ is a generic term, and can include a very wide r of concentration and types of alloying element, whereas pure metals have, by defin only one chemistry.

Master your semester with Scribd  Read Free For 30this Days up to vote on title 3.14 Rank the following in order of increasing Sign thermal conductivity: alumin  & The Newcop-per, York Times useful  Useful  Not silicon, titanium, ceramics, and plastics. Comment on t Special offer for students: Only $4.99/month. applications vis-à-vis these materials.

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Thermal conductivity data is contained in Table 3.1 on p. 89. These materials, in ord

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3.17 What 3.17 What material material properties properties are are desirable desirable for heat shields shields such as those place on the space shuttle?

Material properties properties required for heat shields s hields are sufficient strength so that they do no upon takeoff, reentry, and landing; landing; they must have a high melting point so that they d change phase or degrade at the high temperatures developed during reentry, and must be exceptionally high thermal insulators so that the shuttle cabin does not significantly during reentry.

3.18 List examples of products where materials that are transparent are desired List applications for opaque materials.

This is an open-ended problem, and students should be encouraged to develop their examples based on their insights and experiences. The following are examples of prod where transparency is desired: windows and windshields, bottles, fluid containers (to direct observation of content volumes), wrapping and packaging, and glasses (eyewear) following are examples of products where opacity is desired: windows in restrooms (if pres glass in light bulbs to produce a diffuse light, food packaging to protect the contents from radiation and associated degradation, and product housings for aesthetic reasons.

3.19 Refer to Fig. 3.2 and explain why the trends seen are to be expected.

This is an open-ended problem and students should be encouraged to develop their ob-stervations. The trends are not too surprising qualitatively, but the quantitative na of the trends is at first very surprising. For example, it is not surprising that high-mod  graphite outperforms steel, as people are exposedRead to sporting Free Forequipment 30this Days Sign up to vote on title such as te  racquets that are made of the former but never the people don’t don ’t e Useful However, Not useful  latter.  Cancel anytime. graphite to be 14 times better than steel. Another surprise in the trends is the Special offer for students: Only $4.99/month. performance of glass fibers in an epoxy matrix. However, glass is pretty dense, so w


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3

area under the curve in Fig. 2.6 on p. 63 is about u = = 72, 000 in-lb/in . Assume a density o 3

ρ

◦

= 0.3 lb/in and a specific heat c = = 0.12 BTU/lb F. Therefore, 72, 000 ∆T = = (0.3)(0.12)(778)(12) = 214 F ◦

For (b) 304 stainless steel, we have u = = 175, 000, ρ = = 0.3 and c = = 0.12, hence ∆T = = 520 3

For (c) 1100-H14 aluminum, we have u = = 25, 000 in.-lb/in , ρ = = 0.0975 and c = = 0.215 ◦

hence hence ∆T = = 128 F. 3.21 The natural frequency, f , of a cantilever beam is given by f = 0.56

. EIg 4

wL

where E is is the modulus of elasticity, I is is the moment of inertia, g is is the gra tional constant, w is is the weight of the beam per unit length, and L is is the le of the beam. How does the natural frequency of the beam change, if at al its temperature is increased? Assume that the material is steel.

Let’s Let’s assume that the beam has a square cross section with a side of length h however, that any cross section will result in the same trends, so students shouldn discouraged from considering, for example, circular cross sections. The moment of in for a square cross section is

h Master your semester with Scribd  Free Foron 30this Days Sign up to vote title I = 12 Read  & The NewThe York Times Useful  Not useful moment of inertia will increase as temperature increases, because the cross section 4

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become larger larger due to thermal thermal expansion. expansion. The weight weight per length, w , is given by Special offer for students: Only $4.99/month. W

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h 2 = h 1 (1 + α ∆T )


Read Free Foron 30this Days Sign up to vote title



 Not useful Cancel anytime.

Useful

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L 2 = L 1 (1 + α ∆T )

Therefore, the frequency ratio is . . 4 1 E 1h L f

4

3

3

=

1

1

4

4

.

(1 + α ∆T ) )

E 1hL

=

2

3

E

1

=

1

4

3

E 2h 2 L 1 E 2h 1 (1 + α ∆T ) L 1 f 2 E 2 (1 + α ∆T To compare these effects, consider the case of carbon steel. Figure 2.7 on p. 64 shows a drop in elastic modulus from 190 to 130 GPa over a temperature increase of 1000 C. F Table 3.1 on p. 89, the coefficient coefficient of thermal expansion for steel is 14.5 µm/m C (avera the extreme values given in the table), so that the change in frequency is: ◦

◦

. 1

=

. E

1

=

190

= 1.20

−6 130 [1 + (14 .5 × 10 ) (1000)] E 2 (1 + α ∆T ) Thus, the natural frequency of the beam decreases when heated. This is a general t (and not just for carbon steel), namely that the thermal changes in elastic modulus pl larger role than the thermal expansion of the beam. 2

3.22 It can be shown that thermal distortion in precision devices is low for high va of thermal conductivity divided by the thermal expansion coefficient. Rank materials in Table 3.1 according to their ability to resist thermal distortion. distortion.

Thesemester calculations using the data in Table 3.1 on p. 89 are as follows. When a range Master your with Scribd  of v Read Free For 30 Days Sign up to vote on this title is given for an alloy, the average value has been used. These materials have been  ran & The Newaccording York Times useful to the ratio of thermal conductivity to thermal expansion  Useful  Notcoefficient. Special offer for students: Only $4.99/month.

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Material

Thermal conductivity, k

Thermal expansion coefficient,

k/α

α

Tungsten Molybdenum alloys Copper Silver Silicon Beryllium Gold Copper alloys Aluminum Tantalum alloys Aluminum alloys alloys Columbium (niobium) Nickel Iron Magnesium Magnesium alloys Nickel alloys Steels Titanium Lead alloys Lead Special offer for students: Only $4.99/month. Titanium alloys


166 36.9 4.5 142 5.1 27.8 393 16.5 23.8 429 22.2 19.3 148 7.63 19.4 146 8.5 17.2 317 16.4 19.3 234 18 13.0 222 23.6 9.41 8.31 54 6.5 23 7.72 180 52 7.1 7.32 92 13.3 6.92 74 11.5 6.43 154 26 5.92 106 26 4.10 37 2.42 15.5 14.5 2.28 33 17 8.35 2.04 Read Free Foron 30this Days Sign up to vote title 35 1.20 29.1  Not useful 35  Useful 29.4 1.19 Cancel anytime. 8.8 10 1.14

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Density (kg/m ) Material Aluminum 2700 2630-2820 Aluminum alloys alloys Beryllium 1854 8580 Columbium (niobium) Copper 8970 Copper alloys 7470-8940 Gold 19,300 Iron 7860 Steels 6920-9130 Lead 11,350 Lead alloys 8850-11,350 Magnesium 1745 Magnesium alloys 1770-1780 Molybdenum alloys 10,210 Nickel 8910 Nickel alloys 7750-8850 Silicon 2330 Silver 10,500 Tantalum alloys 16,600 Titanium 4510 Titanium alloys 4430-4700 Tungsten 19,290 Zinc 7140 Special offer for students: Only $4.99/month. 6640-7200 Zinc alloys


Volumetric heat heat Relative volumetric volumetric capacity (J/cm K) (J/kg K) heat capacity 900 2.43 0.58 880-920 2.31-2.59 0.55-0.62 0.83 3.49 1884 272 0.56 2.33 0.82 385 3.45 2.81-3.89 377-435 0.67-0.93 2.49 0.59 129 460 0.86 3.61 448-502 3.10-4.58 0.74-1.09 130 1.47 0.35 1.12-2.13 126-188 0.27-0.51 1.79 1025 0.43 1046 0.44 1.85 0.68 276 2.83 440 3.92 0.94 381-544 0.70-1.15 2.95-4.81 712 1.66 0.40 2.47 0.59 235 2.36 0.56 142 519 2.34 0.56  2.22-2.56 502-544Read Free Foron 30this Days Sign up to vote title 0.53-0.61  138 2.66Not useful 0.64  Useful  Cancel2.75 anytime. 385 0.66 402 2.67-2.89 0.64-0.69 Specific heat

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◦

conductivity = 2 W/mK, coefficient coefficient of thermal expansion =50 µm/m C, electrical resistivity = 182 kΩ-m. kΩ -m. 3

◦

(d) Sugar: density=1587 kg/m , melting point = 185 C, specific heat = 1250 J/kg K, 3

◦

(e) Lithium: density = 535 kg/m , melting point = 180 , specific heat = 3582 J/kg K, 9

−

thermal conductivity = 84.8 W/m K, electrical resistivity = 92.8 × 10 3

◦

Ω-m.

(f) Platinum: density = 21.4 g/cm , melting point = 1768 C, specific heat = 135 J/ thermal conductivity = 70 W/m K, coeffcient of thermal expansion = 19 µm/m ◦

9

−

C, electrical resistivity = 105 × 10

Ω-m.

SYNTHESIS, DESIGN AND PROJECTS 3.25 From your own experience, make a list of parts, components, or products have corroded and have had to be replaced or discarded.

By the student. This is an open-ended problem that have many possible answers, and will vary depending on the background of the student. There are many parts, associ-ated with rusted steel, e.g., automobile frames and bodies, bolts, bicycle pedals Other parts that are commonly corroded include automotive battery cable term marine parts of all kinds (especially if ocean going), nameplates on old machinery, e one extends the discussion to corrosion-assisted failure, one can include just abou parts which fail by fatigue, including shafts, and airplane fuselages as shown below. photograph is a dra-matic example of corrosion-assisted fatigue of an aircraftfuselage Read Free Foron 30this Days Sign up to vote title  Elem occurred mid-flight. ( Source: From Hamrock, B.J., et al., Fundamentals of Machine Useful  Not useful  Cancel anytime. 2nd ed., New York, McGraw-Hill, 2005, p. 265.).


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By the student. This is an open-ended open-ended problem, and many possible answers exist. Som examples are:

(a) High density: Adding weight to a part (like an anchor, bar bells or a boat), a inertial element in a self-winding watch, and weights for vertically sliding wind Also, projectiles such as bullets and shotgun particles are applications where density is advantageous. advantageous.

(b) Low density: Airplane components, aluminum tubing for tents, ladders, and h speed machinery elements. Most sporting goods give better performance if de and hence weight is low, such as tennis rackets, skis, etc.

(c) High melting point: Creep-resistant materials such as for gas-turbine blades or insulation. Mold materials for die casting need to have high melting points, a filaments for light bulbs.

(d) Low melting point: Soldering wire, fuse elements, wax for investment casting, and bricants that depend depend on a phase phase change are examples examples of such applications. applications.

(e) High thermal conductivity: Rapid extraction of heat in radiators and heat exchan and cooling fins for electrical circuits and transformers. Cutting tools with thermal conductivity can help keep temperatures low in machining. Dies in inje molding with high thermal conductivity can extract heat more quickly allowing h production rates.

(f) Low thermal conductivity: Coffee cups, winter clothing, and oven insulation req low thermal conductivity. In addition, handles on cookware, lubricants for forging, and thermos materials (unless evacuated) need low thermal conductivitie  Read Free Foron 30this Days Sign up to vote title  3.27 Describe several applications in which both speciftc strength and speciftc Not useful  Useful  Cancel anytime. stiffness are important.


By the student. This problem is open-ended and the students should be encourage

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Material

Y

(MPa) Aluminum 35 Al alloys 550 Copper 76 Cu alloys 1100 Iron 205 Steels 1725 Lead 14 Pb alloys 14 Magnesium 130 Mg alloys 305 Mo alloys 2070 Nickel 105 Ni alloys 1200 Titanium 344 Ti alloys 1380 Tungsten 550

E

(GPa) 69 79 105 150 190 200 14 14 41 45 360 180 214 80

130 350

Density Spec. strength (kg/m ) (m ×10 ) 2700 1.3 2630 21.3 8970 0.86 7470 15.0 7860 2.66 6920 25.4 0.13 11,350 8850 0.161 7.6 1745 1770 17.6 20.7 10,210 8910 1.2 7750 15.8 4510 7.8 4430 31.7 2.9 19,290

Spec. stiffness (m ×10 ) 2.6 3.1 1.2 2.05 2.5 2.9 0.126 0.16 2.4 2.6 3.6 2.06 2.8 1.8 3.0 1.8

3.30 The maximum compressive force that a lightweight column can withstand be buckling depends on the ratio of the square root of the stiffness to the density the material. For the materials listed in Table 2.2, determine (a) the rati tensile strength to density and (b) the ratio of elastic modulus to den  Comment on the suitability of each for being made into lightweight Read Free Foron 30this Days Sign up to vote title columns.

Master your semester with Scribd  problem uses the results from Problem 3.29. To column, one h & The NewThis York Times Usefula lightweight make  Not useful Cancel anytime.

maximize the specific strength and the specific stiffness. Reviewing the values obtained Special offer for students: Only $4.99/month. can observe that: (a) Pure metals are not useful whereas alloys are much more preferable;

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3.33 Design an actuator to turn on a switch when the temperature drops belo certain level. Use two materials with different coefficients of ther expansion in your design.

This is an open-ended problem with a large number of acceptable answers. The the expansion effect can be used to deform a cantilever, for example, to actuate a switch. A natively, two materials that are long and thin can be welded at their ends. They can the wrapped around around a mandrel, so that after elastic recovery recovery they take on the shape of a s The angular displacement of the ends varies with temperature; a peg attached to one while fixing the other will turn on a switch as soon as the peg peg translates to another another peg retaining fixture. This principal was used to great success in carburators in automo before the 1980s in order to achieve proper proper air/fuel air/fuel ratios as functions functions of temperature temperature


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Download Solutions Manual for Manufacturing Engineering Technology 6th Edition by Serope Kalpakjian and Steven Schmid

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