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Unformatted text preview: 2.19 2.31 If a material does not have an endurance limit (for example, aluminum), how would you estimate its fatigue life? . . . . f
Materials Without endurance limits have their fatigue hfe deﬁnedtiils adcgztaiiginnsrlxgzirfgr
' ’ tress level. For engineering purposes, is e 1 '
cycles to failure at a given s . 11 taken out of servme
' ‘ ' ' art. The part is then usua y estimate of the expected hfetune of a p . I ' ll):fore its lifetime is reached. An alternative approach is to use nondestructive test techniqste:
(Section 36.10 on p. 1132) to periodically measure the accumulated damage 111 a pa , then use fracture mechanics approaches to estimate the remaining life. A paper clip is made of wire 1 mm in diameter. If the original material from
which the wire is made is a rod 18 mm in diameter, calculate the longitudinal
engineering and true strains that the wire has undergone during processing. Engineering strain is deﬁned by Eq. (2.2) on p. 67. Thus, because of volume constancy in
plastic deformation, we may write
2 2
b: = 22 E z 324
10 (if 1.0 Also, letting 10 be unity, the engineering strain is (324 — 1)/1 = 323. True strain is deﬁned by Eq. (2.7) on p. 70. Hence I ,
it 3“ 3578 Note the large difference in strains, even though they both describe the same phenomenon.
(See also the last paragraph in Sec. 2.2.3 on p. 70.) e: 2.32 A strip of metal is 250 mm long. It is stretched in two steps, ﬁrst to 300 mm and then to 400 mm. Show that the total true strain is the sum of the true strains
in each step  that is, that true strains are additive. Show that, in the case of
engineering strains, the strains cannot be added to obtain the total strain. We first calculate the true strains for each step: Step 1: £1 = ln(300/250) = 0.182 Step 2: £2 = 1n(400/300) = 0.288 Thus the total true strain is
6mm; 2 0.182 + 0.288: 0.470 The total strain may also be calculated from the ﬁnal and initial dimensions as
6mm = ln(400/250) = 0.470 thus giving the same answer.
Engineering strains are calculated likewise. Thus, Step 1: e1 2 (300 — 250)/25n = 0200 Step 2: .22 = (400 ‘ 300)/300 = 0.333 Thus the total engineering strain is 0.200+0.333=0.533. However, when calculated from the
initial and ﬁnal dimensions, the total engineering strain is em, = (400 ‘ 250)/250 = 0.600 which is higher than the sum of the engineering strains. Consequently, engineering strains
are not additive, whereas true strains are. ~ 2.33 Identify the two materials in Fig. 2.6 that have the lowest and the highest uniform
elongations. Calculate these quantities as percentages of the original gage lengths. The magnitude of uniform elongation is directly related to the true strain at the onset of
necking. As we see in Fig. 2.5c on p. 71, the necking strain on a true stresstrue strain curve
corresponds to the beginning of the straight portion of the curve. Thus, from Fig. 2.6 on
p. 72 we note that the lowest uniform elongation is for 1112 coldrolled steel, with a necking
strain of about 0.05. The highest is for 304 stainless steel, although 7030 annealed brass is
close to it, with a necking strain of about 0.45. To relate these values to a percentage of the gage length (lo), we convert true strain to percent elongation as follows:
6 = Ina/la) Or, l/lo = e‘. Hence,
Elongation = (e‘ — 1) x 100 Thus, for 1112 coldrolled steel
Elongation = (3°05 — l) x 100 = 5%
And for 7030 brass, the speciﬁc value is Elongation = (ems — l) x 100 = 57% 2.39
as follows: For material A: K = 70,000 psi. n = 0.5, A0 = 0.6 in2
For material B: K = 25,000 psi, n = 0.5, A0 = 0.3 in2 Calculate the maximum tensile force that this cable can withsth prior to neck mg.
Note that necking will occur when 6 = n = 0.5. At this point, the true stresses in each cable
are (using a 2 K6”), respectively, 0,4 = 70,moe°5 = 49,500 psi and
05 = 25,000505 = 17, 700 psi The true areas at necking can be calculated as follows: A A = 0.66”0'5 = 0.364 m2 and _
Ag 2 0.3.245 = 0.182 ml Thus the total load that the cable can support is
Pm“; = (49, 500)(0.364) + (17, 700)(0.182) = 21,200 lb ' 2.50 A Rockwell A test was conducted on a material and a penetration depth of 0.1 mm
was recorded. What is the hardness of the material? What material will typically
give such hardness values? If a Brinell hardness test were to be conducted on this
material, give an estimate of the indentation diameter if the load used is 1500 kg. Horn Fig. 2.12, the Rockwell hardness is calculated from
HRA = 100 — 500t = 100  500(0.1) = 50 HRA me Fig. 2.14 on p. 84, a material with s 50 HRA should have a Brinell hardness of 140
RE. Therefore, from Fig. 2.12, 2P
H3 = (up) (D  MAD—2:35) or. solving for d, 21’ _ _ __2_Q§.0L _
d: D2”(D‘(TDWIF))' "’2 (1° («nonuzsml'3'6m 2.52 List and explain the desirable mechanical properties for (a) an elevator cable, (b)
a paper clip, (c) a leaf spring for a truck, (d) a bracket for a bookshelf, (e) piano
wire, (f) a wire coat hanger, (g) a gasturbine blade. and (h) a staple. (a) Elevator cable: The cable should not elongate elastically to a. great extent or undergo
yielding under the load applied. These requirements call for a material with a high elastic
modulus and yield stress. The cable should also be sufﬁciently flexible and ductile to be
wrapped around the drums during its use. (b) Paper clip: The clip material must possess high ductility to allow it to be formed without
fracture. In its normal use, the clip should recover elastically when removed as well as
apply sufficient force to hold papers together (thus should possess high elastic modulus). (c) Leaf spring: The function of the leaf spring is to absorb energy elastically upon static
or dynamic loading. The energy should be absorbed in an elastic manner because after
absorbing the energy, the spring should return to its original shape. This requires a high
yield stress and high elastic modulus that maximizes the area under the elastic portion
of the stressstrain curve (modulus of resilience). (d) Bracket: Much like an elevator cable. but not as critical, a material with a high yield
stress and elastic modulus would be required. These propertlﬁl ,would keep the shelf
from sagging excessively under load. (6) Piano wire: A piano wire is under high tension (hence high stress) to achieve the desired
tone. The wire should be able to maintain this stress level for a period of time. or
the piano would become out of tune. Thus, its yield stress should be high in order to
sustain the tension required to produce the tone, and should not be susceptible to stress
relaxation. Furthermore, it should be sufﬁciently ductile to be able to be wound around
the tightening mechanisms. (f) Wire coat hanger: Much like the paper clip, the hanger requires a. large amount of
deformation in its manufacturing, so the material must possess high ductility. The
hanger should also be able to maintain its shape when clothes are hung on it, hence it
should possess sufﬁcient yield strength and elastic modulus. (g) Gas turbine blade: A gas turbine blade is required to operate at high temperatures
(depending on its location or stage in the engine), so it should have hightemperature
strength and be resistant to creep, as well as oxidation and corrosion due to combustion products during use. (11) Staple: The properties should closely parallel a paper clip. The staple should have high
ductility to allow it to be deformed without fracture, and also a low yield stress so the
staple can be bent (as well as removed) easily without requiring the user to exert a great
amount of force. 3.15 curves be the unit work done, calculate the temperature rise for (i) 8650 steel,
(ii) 304 stainless steel, and (iii) 1100H14 aluminum. We use the following information given in Chapters 2 and 3: The area under the true stress—
true strain curve and the physical properties for each of the three metals. We then follow
the procedure discussed on pp. 96~97 and use Eq. (2.15). Thus, for (a) 8650 steel, the area
under the curve in Fig. 2.6 on p. 72 is about n == 72, 000 in—lb/ i113. Let’s assume a density of
p = 0.3 lb/in3 and a speciﬁc heat c = 0.12 BTU/1h °F. Therefore, We have 72,000 0 M = (0.3)(0.12)(773)(12) = 214 F For (b) 304 stainless steel, we have u = 175,000, p = 0.3 and c = 0.12, hence AT = 520°F.
For (c) 11001114 aluminum, we have u = 25,000 in.lb/in3, p = 0.0975 and c = 0.215; hence AT = 128°F. 3.17 It can be shown that thermal distortion in precision devices is low for high val
ues of thermal conductivity divided by thermal expansion coefﬁcient. Rank the
materials in Table 3.1 according to their suitability to resist thermal distortion. The calculations using the data in Table 3.1 on p. 103 are as follows. When a range of values
is given for an alloy, the average value has been used. These materials have been ranked
according to the ratio of thermal conductivity to thermal expansion coefficient. Material Thermal Thermal k/a
conductivity, expansion
1: coefﬁcient,
a
Tungsten 166 4.5 36.9
Molybdenum alloys 142 5.1 278
Copper 393 16.5 23.8
Silver 429 19.3 22.2
Silicon 148 7.63 19.4
Beryllium 146 8.5 17.2
Gold 317 19.3 16.4
Copper alloys 234 18 13.0
Aluminum 222 23.6 9.41
Tantalum alloys 54 6.5 8.31
Aluminum alloys 180 23 7.72
Columblum (niobium) 52 7.1 7.32
Nickel 92 13.3 6.92
Iron 74 11.5 6.43
Magnesium 154 26 5.92
Magnesium alloys 106.5 26 4.10
Nickel alloys 37.5 15.5 2.42
Steels 83 14.5 2.28
Titanium 17 8.35 2.04
Lead alloys 35 29.1 1.20 Lead 35 29.4 1.19
Titanium alloys 10 8.8 1.14 \_~ 3.19 List applications where the following properties would be desirable: (am
density, (b) low density, (c) high melting point, ((1) low melting point, (8) high
thermal conductivity, and (f) low thermal conductivity. By the student. This is an open—ended problem, and many possible answers exist. Some
examples are: (a) High density: Adding weight to a part (like an anchor. bar bells or a boat), as an
inertial element in a selfwinding watch. and Weights for vertically sliding windows. Also.
projectiles such as bullets and shotgun particles are applications where high density is
advantageous. (b) Low density: Airplane components, aluminum tubing for tents, ladders, and highsped!
machinery elements. Moat sporting goods give better performance if density and hence ' weight is low, such as tennis rackets, skis. etc. (0) High melting point: Creepresistant materials such m. for gasturbine blades or oven
insulation. Mold materials for die casting need to have high melting points. as do
ﬁlaments for light bulbs. (d) Low melting point: Soldering wire, fuse elements, wax for investment casting, and In»
bricants that depend on a phase change are examples of such applications. (e) High thermal conductivity: Rapid extraction of heat in radiators and heat exclmngcrs,
and cooling fins for electrical circuits and transformers. Cutting tools with high thermal
conductivity can help keep temperatures low in machining. Dies in injection molding
with high thermal conductivity can extract heat more quickly allowing higher production
rates. (f) Low thermal conductivity: Coffee cups, winter clothing, and oven insulation require low
thermal conductivity. In addition, handles on cookware, lubricants for hot forging, and
thermos materials (unless evacuated) need low thermal conductivities. 3.22 For the materials listed in Table 3.1, determine the speciﬁc strength and speciﬁc
stiffness. Describe your observations. r Selected results are as follows (the values which give highest possible quantities have been
used, e.g., high stiﬁ'ness and low density). Data is taken from Table 2.2 on p. 67. Material Y E Density Spec. strength Spec. stiffness
(MP3) (GPa) (kg/m3) (m x103) (m x10“) Aluminum 35 69 2700 1.3 2.6
Al alloys 550 79 2630 21.3 3.1
Copper 76 105 8970 0.86 1.2
Cu alloys 1100 150 7470 15.0 2.05
Iron 205 190 7860 2.66 2.5
Steels 1725 200 6920 25.4 2.9
Lead 14 14 11,350 0.13 0.126
Pb alloys 14 14 8850 0.161 0.16
Magnesium 130 41 1745 ‘ 7.6 2.4
Mg alloys 305 45 1770 17.6 2.6
Mo alloys 2070 360 , 10,210 \ 20.7 3.6
Nickel 105 V 180 ' 8910 1.2 2.06
Ni alloys 1200 214 7750 15.8 2.8
Titanium 344 80 4510 7.8 1.8 Ti alloys 1380 130 4430 31.7 3.0 Tungsten 550 350 19,290 2.9 1.8 ' 36.33 The averageofaverages of a number of samples of size 8 was determined to
be 124. The average range was 17.82 and the standard deviation was 4. The following measurements were taken in a sample: 120, 132, 124, 130, 118, 132,
135, and 121. Is the process in control? Note that for this case the average is ‘ _ 120+132+124+13D+118+132+135+121
m: M =126.5 8 and the range is R = 15. The upper and lower control limits on averages are obtained from
Eqs. (36.6) and (36.7) on B. 1g; as M m
g _ m t .
n 10L; : x r M: = ll‘f‘fo.}mll~h=l50.6
fa; p; f)? ,A,§ 4).;[x’MBitﬂXlnl7Jf
If the sample size is 8, then D4 = 1.864 and D3 = 0.136 (from Table 36.3 on p. 1126). Thus,
from Eqs. (36.8) and (36.9) on p. 1126: UCLR = 04!? = (1.864)(l7.82) = 33.2 LCLR = D31": = (0.136)(17.82) = 2.42 Since both the average and the range of the sample are within the limits, the process is in control. 36.43 Identify the nondestructive techniques that are capable of detecting internal flaws
and those that detect external ﬂaws only. By the student. Typically, nondestructive techniques that detect internal ﬂaws include ul
trasonic, acoustic, radiographic, eddycurrent inspection, thermal impaction, and acoustic
holography. External ﬂaws can be detected by liquid penetrants, magneticparticle inspec—
tion, eddycurrent inspection. thermal inspection, holography, and radiography. ...
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