Page 276 - Materials Science and Engineering An Introduction

248 • Chapter 7 / Dislocations and Strengthening Mechanisms

(b) On the basis of these resolved shear stress 7.26 If it is assumed that the plot in Figure 7.15 is for
values, which slip system(s) is (are) most favora- non-cold-worked brass, determine the grain size
bly oriented? of the alloy in Figure 7.19; assume its composition
7.17 Consider a single crystal of some hypothetical is the same as the alloy in Figure 7.15.
metal that has the BCC crystal structure and
is oriented such that a tensile stress is applied Solid-Solution Strengthening
along a [121] direction. If slip occurs on a (101) 7.27 In the manner of Figures 7.17b and 7.18b,
plane and in a [111] direction, compute the stress indicate the location in the vicinity of an edge
at which the crystal yields if its critical resolved dislocation at which an interstitial impurity atom
shear stress is 2.4 MPa. would be expected to be situated. Now, briefly
explain in terms of lattice strains why it would be
7.18 Consider a single crystal of some hypothetical situated at this position.
metal that has the FCC crystal structure and is
oriented such that a tensile stress is applied along
a [112] direction. If slip occurs on a (111) plane Strain Hardening
and in a [011] direction, and the crystal yields at a 7.28 (a) Show, for a tensile test, that
stress of 5.12 MPa, compute the critical resolved
shear stress. %CW = a P b * 100
P + 1
7.19 The critical resolved shear stress for copper
(Cu) is 0.48 MPa (70 psi). Determine the maxi- if there is no change in specimen volume during
mum possible yield strength for a single crystal the deformation process (i.e., A 0 l 0 A d l d ).
of Cu pulled in tension. (b) Using the result of part (a), compute
the percent cold work experienced by naval
Deformation by Twinning brass (for which the stress–strain behavior is
7.20 List four major differences between deforma- shown in Figure 6.12) when a stress of 415 MPa
tion by twinning and deformation by slip relative (60,000 psi) is applied.
to mechanism, conditions of occurrence, and Two previously undeformed cylindrical speci-
final result. 7.29
mens of an alloy are to be strain hardened
by reducing their cross-sectional areas (while
Strengthening by Grain Size Reduction maintaining their circular cross sections). For
7.21 Briefly explain why small-angle grain bounda- one specimen, the initial and deformed radii are
ries are not as effective in interfering with the slip 15 and 12 mm, respectively. The second speci-
process as are high-angle grain boundaries. men, with an initial radius of 11 mm, must have
7.22 Briefly explain why HCP metals are typically the same deformed hardness as the first speci-
more brittle than FCC and BCC metals. men; compute the second specimen’s radius after
deformation.
7.23 Describe in your own words the three strengthen-
ing mechanisms discussed in this chapter (i.e., grain 7.30 Two previously undeformed specimens of the
size reduction, solid-solution strengthening, and same metal are to be plastically deformed by
strain hardening). Explain how dislocations are reducing their cross-sectional areas. One has a
involved in each of the strengthening techniques. circular cross section, and the other is rectangu-
lar; during deformation, the circular cross section
7.24 (a) From the plot of yield strength versus (grain
diameter) 1/2 for a 70 Cu–30 Zn cartridge brass in is to remain circular, and the rectangular is to
Figure 7.15, determine values for the constants s 0 remain rectangular. Their original and deformed
and k y in Equation 7.7. dimensions are as follows:
(b) Now predict the yield strength of this Circular Rectangular
alloy when the average grain diameter is (diameter, mm) (mm)
3
2.0 10 mm. Original
7.25 The lower yield point for an iron that has an dimensions 18.0 20 50
average grain diameter of 1 10 2 mm is 230 Deformed
3
MPa (33,000 psi). At a grain diameter of 6 10 dimensions 15.9 13.7 55.1
mm, the yield point increases to 275 MPa (40,000
psi). At what grain diameter will the lower yield Which of these specimens will be the hardest
point be 310 MPa (45,000 psi)? after plastic deformation, and why?