7.9
Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crystal structures,
are of the form
b=
a
uvw
2
where a is the unit cell edge length. Also, since the magnitudes of these Burgers vectors may be
determined from the following equation:
b =
a 2 2
(u + v + w2 )1/2
2
(7.10)
determine values of |b| for copper and molybdenum. You may want to consult Table 3.1.
Solution
For Cu, which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and a = 2 R 2 =
0.3614 nm (Equation 3.1); also, from Equation 7.1a, the Burgers vector for FCC metals is
b =
a
⟨110⟩
2
Therefore, the values for u, v, and w in Equation 7.10 are 1, 1, and 0, respectively. Hence, the
magnitude of the Burgers vector for Cu is
b =
=
a
2
u 2 + v 2 + w2
0.3614 nm
(1) 2 + (1) 2 + (0) 2 = 0.2555 nm
2
For Mo which has a BCC crystal structure, R = 0.1363 nm (Table 3.1) and a =
4R
3
= 0.3147
nm (Equation-3.3); also, from Equation 7.1b, the Burgers vector for BCC metals is
b =
a
⟨111⟩
2
Therefore, the values for u, v, and w in Equation 7.10 are 1, 1, and 1, respectively. Hence, the
magnitude of the Burgers vector for Mo is
b =
0.3147 nm
(1) 2 + (1) 2 + (1)2 = 0.2725 nm
2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
7.12 Consider a metal single crystal oriented such that the normal to the slip plane and the slip
direction are at angles of 43.1° and 47.9°, respectively, with the tensile axis. If the critical
resolved shear stress is 22 MPa, will an applied stress of 50 MPa cause the single crystal to
yield? If not, what stress will be necessary?
Solution
This problem calls for us to determine whether or not a metal single crystal having a specific
orientation and of given critical resolved shear stress will yield. We are given that φ = 43.1°, λ =
47.9°, and that the values of the critical resolved shear stress and applied tensile stress are 22 MPa and
50 MPa respectively. From Equation 7.2
τ R = σ cos φ cos λ = (50 MPa)(cos 43.1°)(cos 47.9°) = 24.5 MPa
Since the resolved shear stress (24.5 MPa) is greater than the critical resolved shear stress (22 MPa),
the single crystal will yield.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
7.17
Consider a single crystal of some hypothetical metal that has the FCC crystal structure and
is oriented such that a tensile stress is applied along a [102] direction. If slip occurs on a (111)
plane and in a [101] direction, compute the stress at which the crystal yields if its critical
resolved shear stress is 3.2 MPa.
Solution



 (u12 + v12 + w12 )(u22 + v22 + w22 ) 

(−1)(−1) + (0)(0) + (2)(1)


2
2
2
(−1) + 0 + 2  (−1) + 0 + 1 

3 
=
 18.4°
10 
λ = cos −1 


= cos 


= cos −1 

−1
φ = cos
−1

= cos −1 

u1u2 + v1v2 + w1w2


 (−1)(1) + (0)(1) + (2)(1) 


2
2
 (−1) + 0 + 2  [1 + 1 + 1] 
1 
=
 75.03°
15 
τ crss
3.2 Mpa
=
cos φ cos λ  3   1 



 10   15 
= 13.06 Mpa
=
σy
7.23 (a) From the plot of yield strength versus (grain diameter)–1/2 for a 70 Cu–30 Zn cartridge brass,
Figure 7.15, determine values for the constants σ0 and ky in Equation 7.7.
(b) Now predict the yield strength of this alloy when the average grain diameter is 2.0 × 10−3
mm.
Solution
(a) Perhaps the easiest way to solve for σ0 and ky in Equation 7.7 is to pick two values each of σy and
d-1/2 from Figure 7.15, and then solve two simultaneous equations, which may be created. For example
d−1/2 (mm)−1/2
σy (MPa)
4
75
12
175
The two equations are thus
75 = σ 0 + 4 k y
175 = σ 0 + 12 k y
Solution of these equations yield the values of
k y = 12.5 MPa (mm)1/2
σ0 = 25 MPa
(b) When d = 2.0 × 10−3 mm, d−1/2 = 22.36 mm−1/2, and, using Equation 7.7,
σ y = σ 0 + k y d −1/2
1/2
= (25 MPa) + ⎡⎢12.5 MPa (mm) ⎤⎥ (22.36 mm −1/2 ) = 305 MPa
⎣
⎦
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
Strain Hardening
7.27
(a) Show, for a tensile test, that
⎛ ∈ ⎞
%CW = ⎜
⎟ × 100
⎝ ∈ +1 ⎠
if there is no change in specimen volume during the deformation process (i.e., A0l0 = Adld).
(b) Using the result of part (a), compute the percent cold work experienced by naval brass (the
stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa is applied.
Solution
(a) From Equation 7.8
⎛ A − Ad ⎞
⎛
Ad ⎞
%CW = ⎜ 0
⎟ × 100 = ⎜ 1 −
⎟ × 100
A0 ⎠
⎝ A0 ⎠
⎝
Which is also equal to
⎛
l ⎞
⎜1 − 0 ⎟ × 100
ld ⎠
⎝
since Ad/A0 = l0/ld, the conservation of volume stipulation given in the problem statement. Now, from
the definition of engineering strain (Equation 6.2)
∈=
ld − l0
l
= d −1
l0
l0
Or,
l0
1
=
ld ∈ + 1
Substitution for l0/ld into the %CW expression above gives
⎛
⎛
⎛ ∈ ⎞
l ⎞
1 ⎞
%CW = ⎜ 1 − 0 ⎟ × 100 = ⎜1 −
⎟ × 100 = ⎜
⎟ × 100
∈ + 1⎠
ld ⎠
⎝
⎝∈ + 1 ⎠
⎝
(b) From Figure 6.12, a stress of 400 MPa corresponds to a strain of 0.13. Using the above expression
⎛ ∈ ⎞
⎛
⎞
0.13
%CW = ⎜
⎟ × 100 = ⎜
⎟ × 100 = 11.5%CW
⎝∈ + 1 ⎠
⎝ 0.13 + 1.00 ⎠
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
7.29 Two previously undeformed specimens of the same metal are to be plastically deformed by
reducing their cross-sectional areas.
One has a circular cross section, and the other is
rectangular; during deformation the circular cross section is to remain circular, and the
rectangular is to remain as such. Their original and deformed dimensions are as follows:
Circular
Rectangular
(diameter, mm)
(mm)
Original dimensions
15.2
125 × 175
Deformed dimensions
11.4
75 × 200
Which of these specimens will be the hardest after plastic deformation, and why?
Solution
The hardest specimen will be the one that has experienced the greatest degree of cold work.
Therefore, all we need do is to compute the %CW for each specimen using Equation 7.8. For the
circular one
⎡A − A ⎤
d
%CW = ⎢ 0
⎥ × 100
A0
⎣
⎦
2
⎡ ⎛ 15.2 mm ⎞ 2
⎛ 11.4 mm ⎞ ⎤
⎢π ⎜
⎟ − π⎜
⎟ ⎥
2
2
⎠
⎝
⎠ ⎥ × 100 = 43.8%CW
=⎢ ⎝
2
⎢
⎥
15.2
mm
⎛
⎞
π⎜
⎢
⎥
⎟
2
⎝
⎠
⎣⎢
⎦⎥
2
⎡ ⎛ 15.2 mm ⎞ 2
⎛ 11.4 mm ⎞ ⎤
π
π
−
⎢ ⎜
⎟
⎜
⎟ ⎥
2
2
⎠
⎝
⎠ ⎥ × 100 = 43.8%CW
=⎢ ⎝
2
⎢
⎥
⎛ 15.2 mm ⎞
π⎜
⎢
⎥
⎟
2
⎝
⎠
⎣⎢
⎦⎥
For the rectangular one
⎡ (125 mm)(175 mm) − (75 mm)(200 mm) ⎤
%CW = ⎢
⎥ × 100 = 31.4%CW
(125 mm)(175 mm)
⎣
⎦
Therefore, the deformed circular specimen will be harder.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
7.38 The average grain diameter for a brass material was measured as a function of time at 650°C,
which is tabulated below at two different times:
Time
Grain Diameter
(min)
(mm)
30
3.9 × 10–2
90
6.6 × 10–2
(a) What was the original grain diameter?
(b) What grain diameter would you predict after 150 min at 650°C?
Solution
(a) Using the data given and Equation 7.9 (taking n = 2), we may set up two simultaneous equations
with d0 and K as unknowns; thus
(3.9 × 10−2 mm)2 − d 02 = (30 min) K
(6.6 × 10−2 mm)2 − d 02 = (90 min) K
Solution of these expressions yields a value for d0, the original grain diameter, of
d0 = 0.01 mm,
−5
and a value for K of 4.73 × 10 mm /min
2
(b) At 150 min, the diameter d is computed using a rearranged form of Equation 7.9 as
d =
=
d02 + Kt
(0.01 mm) 2 + (4.73 × 10−5 mm2 /min) (150 min) = 0.085 mm
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.
7.41 An uncold-worked brass specimen of average grai n size 0.009 mm has a y ield strength of 160
MPa. Estimate the yield strength of this alloy after it has been heated to 600 °C for 1000 s, i f it
is known that the value of ky is 12.0 MPa⋅mm1/2.
Solution
In order to solve this problem, it is first necessary to calculate the constant σ0 in Equation 7.7
as
σ 0 = σ y − k y d −1/ 2
= 160 MPa − (12.0 MPa ⋅ mm1/2 )(0.009 mm) −1/ 2 = 33.5 MPa
Next, we must determine the average grain size af ter the heat treatment. From Figure 7.25 at 600°C
(873 K) after 1000 s (16.7 min) the average grain size of a brass material is about 0.20 mm. Therefore,
calculating σ y at this new grain size using Equation 7.7 we get
σ y = σ 0 + k y d −1/2
= 33.5 MPa + (12.0 MPa ⋅ mm1/2 ) (0.020 mm) −1/2 =60.3 MPa
Excerpts from this wor k may be r eproduced by instructors for distribution on a not- for-profit basis for testing or instructiona l
purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of
this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the
copyright owner is unlawful.