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Unformatted text preview: 284 ' Gupta’8 l DebnnaﬁonandStrengHIeningMedlanisms QUESTIONS AND PROBLEMS Additional problems and questions for this chapter may be found on both Student and
instructor Companion Sites at. m m'ley. com/college/calllster. ——~—————————~—___....____..—_,.__—_____—________—__ Basic Concepts
Characlenlsﬂ'c.‘ of Dislocations 8.1 To provide some perspective on the dimen
sions of atomic defects, consider a metal spec
imen that has a dislocation density of 105 mm2 .Suppose that all the dislocations' 1n 1000
mm3 (1 cm3) were somehow removed and
linked end to end. How far (in miles) would
this chain extend? Now suppose that the den
sity is increased to 109 mm‘2 by cold working.
What would be the chain length of dislocations
in 1000 mm3 of material? 8.2 Consider two edge dislocations of opposite
sign and having slip planes that are separated
by several atomic distances as indicated in
the diagram. Brieﬂy describe the defect that
results when these two dislocations become
aligned with each other. _J_"
HT Slip Systems 8.3 (11) Compare planar densities (Section 3.15
and Problem W146 [which appears on the
book’s Web site]) for the (100), (110), and
(111) planes for FCC. 01) Compare planar densities (Problem 3.44)
for the (100). (110), and (111) planes for
BCC. 8.4 One slip system for the BCC crystal structure
is {110}(111). In a manner similar to Figure
8.6!), sketch a {110}type plane for the BCC
structure, representing atom positions with cir
cles. Now, using arrows, indicate tWo different
(111} slip directions within this plane. 8.5 Equations 8.1a and 8.1 b, expressions for Burg ers vectors for FCC and BCC crystal struc
tures, are of the form I) = gummy) where a is the unit cell edge length. Since the
magnitudes of these Burgers vectors may be ‘ determined from the following equation: lbl = $03 + 112 + 1112f” (3.13)
determine values of lbj for copper and iron. You may want to consult Equations 3.1 and
3.3 as well as Table 3.1.
Slip in Single Crystals 8.6 Consider a metal single crystal oriented so that
the normal to the slip plane and the slip direc
tion are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved
shear stress is 6.2 MPa (900 psi), will an ap 7 plied stress of 12 MPa (1750 psi) cause the sin ' gle crystal to yield? If not, what stress will be I; necessary? 8.? Consider a single crystal of nickel oriented
such that a tensile stress is applied along a [001] direction If slip occurs on a (111) plane and 1a
a [101] direction, and 1s initiated at an applied
tensile stress of 13. 9 MPa (2020 psi), compute
the critical resolved shear stress. 8.8 (a) A single crystal of a metal that has the
BCC crystal structure is oriented so theta tensile stress is applied in the [100] direc— ..
tion. If the magnitude of this stress is 4.0'
MPa compute the resolved shear stress in =
the [111] direction on each of the (110),. (011) and (101) planes. (b) On the basis of these resolved shear stress
values, which slip system(s)' is (are) most
favorably oriented? 8.9 The critical resolved shear stress for copper is
0.48 MPa (70 psi). Determine the maximum
possible yield strength for a single crystal of
Cu pulled 1n tension. Strengthening by Grain Size Reduction
8.10 Brieﬂy explain why smallangle grain bound aries are not as effective in interfering with the _'
slip process as are highangle grain boundaries. ' 8.11 Brieﬂy explain why IlCP metals are typically _ more brittle than FCC and BCC metals. 8.12 (a) From the plot of yield strength versus;
(grain diaineterr“2 for a 70 Cu—30 Zn_ l
g
l cartridge brass (Figure 8.15). determine
values for the constants no and k,, in Equa—
tion 8.7. (11) Now predict the yield strength of this alloy
when the average grain diameter is 2.0 x
10‘3 mm. 8.13 If it is assumed that the plot in Figure 8.15
is for noncoldworked brass, determine the
grain size of the alloy in Figure 8.19; assume
its composition is the same as the alloy in
Figure 8.15. Strain Hardening 8.14 'IKvo previously undeformed cylindrical spec
imens of an alloy are to be strain hardened
by reducing their crosssectional areas (while
maintaining their circular cross sections). For
one specimen, the initial and deformed radii
are 15 mm and 12 mm, respectively. The sec
ond specimen, with an initial radius of 11 mm,
must have the same deformed hardness as the
ﬁrst specimen: compute the second specimen‘s
radius after deformation. 8.15 A cylindrical specimen of cold—worked cop
per has a ductility (%EL) of 15%. If its cold
worked radius is 6.4 mm (0.25 in.), what was
its radius before deformation? 8.16 Experimentally, it has been observed for single
crystals of a number of metals that the critical
resolved shear stress rum is a function of the
dislocation density p0 as Tm: = rGtAJpD where to and A are constants. For copper.
the critical resolved shear stress is 0.69 MPa
(100 psi) at a dislocation density of 104 min—2.
If it is known that the value of to for copper
is 0.069 MPa (10 psi), compute the rm at a
dislocation density of 10‘3 mm‘z. Recovery
Recrystaﬂfzaﬁon
Grain Growth 8.17 Estimate the fraction of recrystallization from
the photomicrograph in Figure 8.21a. 8.18 The average grain diameter for a brass ma
terial was measured as a function of time at
650° C. which is tabulated below at two differ
ent times: Questions and Problems 0 285 Time (min) Grain Diameter (your)
40 5.6 x 10"2
100 8.0 x 10'2 (a) What was the original grain diameter?
(b) What grain diameter would you predict af
ter 200 min at 650°C? 8.19 Grain growth is strongly dependent on tem
perature (i.e., rate of grain growth increases
with increasing temperature), yet temperature
is not explicitly given as a part of Equation
8.9. (a) into which of the parameters in this ex
pression would you expect temperature to
be included? (b) 0n the basis of your intuition. cite an ex
plicit expression for this temperature de
pendence. 8.20 An uncoldworked brass specimen of average
grain size 0.01 mm has a yield strength of 150
MPa (21,750 psi). Estimate the yield strength
of this alloy after it has been heated to 500°C
for 1000 3, if it is known that the value of on is
25 MP3 (3625 psi). Defamation ofSerru'cryrtaMne Polymers
(Deformation of Hortense“) 8.21 In your own words, describe the mechanisms
by which semicrystalline polymers (2) elasti
cally deform and (h) plastically deform. and
(c) by which elastomers elastically deform. Factors That inﬂuence the Mechanical Properties
ofSemicryrtalEne Polymer:
Deformation of Elastornen 3.22 Briefly explain how each of the following in
ﬂuences the tensile or yield strength of a
semicrystalline polymer and why: (a) Molecular weight 0)) Degree of crystallinity (c) Deformation by drawing (d) Annealing of an undeformed material 8.23 The tensile strength and numberaverage
molecular weight for two poly(rnethyl meth
acrylate) materials are as follows: 286 ' Chapter 8 i Defamation and Strengthening Mechanisms Tensile Nam barA vemgc Molecular
Strength (MP0) Weight {g/mol)
50 30,000
150 50,000 Estimate the tensile strength at a number
average molecular weight of 40,000 g/mol. 8.24 For each of the following pairs of polymers,
do the following: (1) state whether or not it is
possible to decide whether one polymer has a
higher tensile modulus than the other; (2) if
this is possible, note which has the higher tenv
sile modulus and then cite the reason(s) for
your choice; and (3) if it is not possible to de
cide, then state why. (a) Branched and atactic poly(vinyl chlo
ride) with a weight~average molecular
weight of 100,000 g/moi; linear and isotac
tic poly(vinyl chloride) having a weight
average molecular weight of 75,000 g/mol (1)) Random styrenebutadiene copolymer
with 5% of possible sites crosslinked;
block styrenebutadiene copolymer with
10% of possible sites crosslinked (c) Branched polyethylene with a number
average molecular weight of 100,000
g/mol; atactic polypropylene with a
numberaverage molecular weight of
150.000 g/mol 8.25 For each of the following pairs of polymers,
plot and label schematic stress—strain curves
on the same graph (i.e., make separate plots
for parts a, b, and c). (a) Polyisoprene having a numberaverage
molecular Weight of 100,000 glmol and
10% of available sites crosslinked; poly~ DESIGN PROBLEMS isoprene having a numberaverage molec
ular weight of 100,000 g/mol and 20% of
available sites crosslinlted (b) Syndiotactic polypropylene having a
Weightaverage molecular weight of
100,000 g/mol; atactic polypropylene hav
ing a weightaverage molecular weight of
75,000 glmol (c) Branched polyethylene having a number
average molecular weight of 90,000 g/Inol;
heavily crosslinked polyethylene having
a numberaverage molecular weight of
90,000 g/mol 826 Which of the following would you expect to be
elastomers and which thermosetting polymers
at room temperature? Justify each choice. (a) Linear and highly crystalline polyethylene (b) Heavily crosslinked polyisoprene having
a glasstransition temperature of 50°C
(122° F) (c) Linear and partially amorphous poly(vinyl
chloride) 8.21!r Fifteen kilogram of polychloroprene is vul canized with 5.2 kg sulfur. What fraction of '
the possible crosslink sites is bonded to sulfur  crosslinks, assuming that, on the average, 5.5
sulfur atoms participate in each crosslink? 8.28 The vulcanization of polyisoprene is accom
plished with sulfur atoms according to Equa
tion 8.12. If 45.3 wt% sulfur is combined with
polyisoprene, how many crosslinks will he as
sociated with each isoprene repeat unit if it is
assumed that, on the average, ﬁve sulfur atoms
participate in each crosslink? 8.29 Demonstrate, in a manner similar to Equation
8.12. how vulcanization may occur in a chloro~
prene rubber. M Strain Hardening
Recrystallization 8.Dl Determine whether or not it is possible to cold work steel so as to give a minimum Brinell
hardness of 240, and at the same time have a ductility of at least 15 %EL. Justify your deci
sion. 8.D2 A cylindrical specimen of cold~worked steel has a Brinell hardness of 240.
(a) Estimate its ductility in percent elonga
tion. (b) If the specimen remained cylindrical dur ing deformation and its original radius 1 was 10 mm (0.40 in), determine its radius .7 after deformation. 8.03 A cylindrical rod of 1040 steel originally 11.4 mm (0.45 in.) in diameter is to be cold worked
by drawing; the circular cross section will
be maintained during defamation. A cold
worked tensile strength in excess of 825 MPa
(120,000 psi) and a ductility of at least 12% EL
are desired. Furthermore. the ﬁnal diameter
must be 8.9 mm (0.35 in.). Explain how this
may be accomplished. Design Problems ' 281 8.D4 A cylindrical brass rod having a minimum ten sile strength of 450 MPa (65,000 psi), a ductil
ity of at least 13% EL, and a ﬁnal diameter of
12.7 mm (0.50 in.) is desired. Some 19.0 mm
(0.75 in.) diameter brass stock that has been
cold worked 35% is available. Describe the
procedure you would follow to obtain this ma
terial. Assume that brass experiences crack
ing at 65 % CW. ...
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 Fall '08
 GRONSKY

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