squares lie within the recrystallized areas, and since there are 672 total
squares, the specimen is about 60% recrystallized. 7.34 During coldworking, the grain structure of the metal has been distorted to
accommodate the deformation. Recrystallization pr

8.32 This problem asks that we determine the maximum lifetimes of
continuous driving that are possible at an average rotational velocity of
600 rpm for the alloy the fatigue data of which is provided in Problem
8.31 and at a variety of stress levels. (a)

From Figure 7.17(b), at least 27%CW is required for a tensile strength of
450 MPa. Furthermore, according to Figure 7.17(c), 13%EL corresponds
a maximum of 30%CW. Let us take the average of these two values (i.e.,
28.5%CW), and determine what previous spe

cost analysis on the above alloys. Below are tabulated the y 1/ 2
values, the relative material cost (c), and the product of these two
parameters; also those alloys for which cost data are provided are
ranked, from least to most expensive.
_ Alloy
Conditi

the three slip systems. In the sketch below is shown the unit cell and
(111)-[11 0 ] slip configuration. 178 The angle between the [100] and [1
1 0] directions, , is 45. For the (111) plane, the angle between its
normal (which is the [111] direction) and

fractured because it crystallized" is erroneous inasmuch as the metal
was crystalline prior to being stressed (virtually all metals are
crystalline). 8.38 (a) With regard to size, beachmarks are normally of
macroscopic dimensions and may be observed with

must determine the average grain size after the heat treatment. From
Figure 7.23 at 500C after 1000 s (16.7 min) the average grain size of a
brass material is about 0.016 mm. Therefore, calculating y at this new
grain size using Equation (7.5) we get y =

same alloy at one stress level and another internal 202 crack length. It
first becomes necessary to solve for the parameter Y for the conditions
under which fracture occurred using Equation (8.5). Therefore, Y = KIc
a = 26 MPa m (112 MPa) () 8.6 x 103 m

of 415 MPa (60,000 psi) corresponds to a strain of 0.16. Using the above
expression %CW = + 1 x 100 = 0.16 0.16 + 1.00
x 100 = 13.8%CW 7.27 In order for these two cylindrical specimens to
have the same deformed hardness, they must be deformed to the same

_ It is up
to the student to select the best metal alloy to be used for this plate on
a strength-per-mass basis, including the element of cost, as well as
other relevant considerations. (f) The student should use his or her own
discretion in the selection

MPa) (0.20 mm) /2 0.001mm = 15 GPa 2.2 x 106 ( psi) Therefore,
fracture will not occur since this value is less than E/10. 8.6 We may
determine the critical stress required for the propagation of an internal
crack in aluminum oxide using Equation (8.3); t

mm (0.45 in.). o ' o ' 7.D6 Let us first calculate the percent cold work
and attendant yield strength and ductility if the drawing is carried out
without interruption. From Equation (7.6) %CW = do 2 2
dd 2 2 do 2 2 x 100 193 = 10.2 mm 2
2 7.6 mm 2 2 1

without exceeding the 65%CW limit. Again employing Equation (7.6)
%CW = 23.6 mm 2 2 15.0 mm 2 2 23.6 mm
2 2 x 100 = 59.6%CW In summary, the procedure which can
be used to produce the desired material would be as follows: cold work
the as-received stock to

(7000 psi). Taking the natural logarithm of Equation (8.20) yields ln s
= ln K2 + n ln Qc RT With the given data there are two unknowns in
this equation-namely K2 and n. Using the data provided in the problem
we can set up two independent equations as fol

pin are used to connect a gear or pulley to a shaft-the pin is designed
shear off before damage is done to either the shaft or gear in an
overload situation. 8.2W The theoretical cohesive strength of a material
is just E/10, where E is the modulus of elas

of these values into the above equation leads to Qc = (8.31 J/mol -K)
ln 107 ( ) ln 0.8 x 105 ( ) 1 700 K 1 811 K =
186,200 J/mol (b) We are now asked to calculate s at 649C (922 K).
It is first necessary to determine the value of K , which is accomplishe

application), , we consult the illustration of this same unit cell as
shown below. For the triangle ABC, the angle is equal to tan1 A B B C
. The length B C is equal to the unit cell edge length a. From
triangle ABD, A B = a 2 , and, therefore, = a 2 a =

2.3 mm (0.09 in.) For the 4340 alloy steel tempered at 425C B = (2.5)
87.4 MPa m 1420 MPa 2 = 0.0095 m = 9.5 mm (0.38 in.) 201
8.14 This problem asks us to determine whether or not the 4340 steel
alloy specimen will fracture when exposed to a stress of 10

creep rate at a stress of 83 MPa (12,000 psi) and 1300 K. Taking the
natural logarithm of Equation (8.20) yields 218 ln s = ln K2 + n ln
Qc RT With the given data there are two unknowns in this equation-namely K2 and Qc. Using the data provided in the pr

an elevated temperature in order to allow recrystallization and some
grain growth to occur until the average grain diameter is 0.020 mm.
7.40 (a) The temperature dependence of grain growth is incorporated
into the constant K in Equation (7.7). (b) The exp

metal alloys in the database that have strength performance indices
greater than 6.0 (in SI units). (Note: for this performance index of 6.0,
density has been taken in terms of g/cm3 rather than in the SI units of
kg/m3.) Twelve alloys satisfy this criter

strains associated with the edge dislocation; such tensile strains exist
just below the bottom of the extra half-plane of atoms (Figure 7.4). 7.25
The hardness measured from an indentation that is positioned very
close to a preexisting indentation will be

Harden the outer surface of the structure by case hardening
(carburizing, nitriding) or shot peening. 8.40 Creep becomes important
at 0.4T m, T m being the absolute melting temperature of the metal. For
Ni, 0.4T m = (0.4)(1455 + 273) = 691 K or 418C (785F

of the most densely packed crystallographic plane, and within that
plane the most closely packed direction. This plane and direction will
vary from crystal structure to crystal structure. 7.6 (a) For the FCC
crystal structure, the planar density for the (

write y2 + x 2 2 = x2 which leads to y = x 3 2 . And,
substitution for the above expression for x yields y = x 3 2 = 4R 2 3
3 2 = 4R 2 2 Thus, the area of this triangle is equal to
AREA = 1 2 x y = 1 2 4R 2 3 4 R 2 2 = 8R2
3 And, finally, the planar den

cos(45)cos(54.7) = (4.0 MPa)(0.707)(0.578) = 1.63 MPa 182 (b) The
most favored slip system(s) is (are) the one(s) that has (have) the largest
value. Both (110)-[1 R 1 1] and (10 1 )-[1 1 1] slip systems are most
favored since they have the same R (1.63 M

100 = 56%CW At 56%CW, the steel will have a tensile strength on the
order of 920 MPa (133,000 psi) [Figure 7.17(b)], which is adequate;
however, the ductility will be less than 10%EL [Figure 7.17(c)], which is
insufficient. Instead of performing the drawi

7.30 (a) We want to compute the ductility of a brass that has a yield
strength of 275 MPa (40,000 psi). In order to solve this problem, it is
necessary to consult Figures 7.17(a) and (c). From Figure 7.17(a), a yield
strength of 275 MPa for brass correspo

plane. Furthermore, the (111) plane does not pass through the center
of atom D, which is located at the unit cell center. The atomic packing of
this plane is presented in the following figure; the corresponding atom
positions from the Figure (a) are also

7.13, and then set up and solve two simultaneous equations. For
example d-1/2 (mm)-1/2 y (MPa) 4 75 12 175 The two equations are
thus 75 = o + 4ky 183 175 = o + 12ky These yield the values of ky =
12.5 MPa(mm)1/2 1810 psi( ) mm 1/ 2 [ ] o = 25 MPa (3630 p

component using the same alloy for another stress level and internal
crack length. It first becomes necessary to solve for the parameter Y for
the conditions under which fracture occurred using Equation (8.5).
Therefore, Y = KIc a = 40 MPa m (300 MPa) ()

500C, the time necessary for the average grain diameter to increase
from 0.01 to 0.1 mm is approximately 3500 min. (b) At 600C the time
required for this same grain size increase is approximately 150 min. 7.38
(a) Using the data given and Equation (7.7) a

MPa (130 psi) 7.13 This problem asks that we compute the critical
resolved shear stress for silver. In order to do this, we must employ
Equation (7.3), but first it is necessary to solve for the angles and
from the sketch below. 177 If the unit cell edge

2.0; furthermore, the r/h ratio is (4 mm)/(20 mm) = 0.20. Using the w/h
= 2.0 curve in Figure 8.2cW, the Kt value at r/h = 0.20 is 1.8. And since Kt
= m o , then m = Kt o = (1.8)(140 MPa) = 252 MPa (36,000 psi) (b)
Now it is necessary to determine how muc

stress at which yielding begins; it is a property of the material. 7.10 We
are asked to compute the Schmid factor for an FCC crystal oriented with
its [100] direction parallel to the loading axis. With this scheme, slip
may occur on the (111) plane and in