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Class Notes
(839,150)

Canada
(511,218)

University of Toronto St. George
(44,098)

MSE101H1
(44)

Scott Ramsay
(44)

Lecture

Department

Materials Science & Engineering

Course Code

MSE101H1

Professor

Scott Ramsay

Description

CHAPTER 8
DEFORMATION AND STRENGTHENING MECHANISMS
PROBLEM SOLUTIONS
Slip Systems
8.3 (a) Compare planar densities (Section 3.15 and Probl em W3.46 [which appears on the book’s Web
site]) for the (100), (110), and (111) planes for FCC.
(b) Compare planar densities (Problem 3.44) for the (100), (110), and (111) planes for BCC.
Solution
(a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.12 as
PD (FCC) = 1 = 0.177
110 4R 2 2 R2
Furthermore, the planar densities of the (1nd (111) planes are calculated in Homework Problem
W3.46, which are as follows:
1 0.25
PD100(FCC) = 2 = 2
4R R
1 0.29
PD111(FCC) = 2 = 2
2R 3 R
(b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in
Homework Problem 3.44, which are as follows:
PD (BCC) = 3 = 0.19
100 16R 2 R 2
3 0.27
PD110(BCC) = 2 = 2
8R 2 R
Excerpts from this work may be reproduced bdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textboAny 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. Below is a BCC unit cell, withinwhich is shown a (111) plane.
(a)
The centers of the three corner atoms, denoted by A, B, and C lie on this plan e. Furthermore, the (111) plane does
not pass through the center of atom D, which is located at th e 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 noted.
(b)
Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count. One sixth of
each of the three atoms labeled A, B, a nd C is associated with this plane, which gives an equivalence of one-half
atom.
In Figure (b) the triangle with A, B, and C at its corners is an equilateral triangle. And, from Figure ( b),
xy
the area of this triangle is . The triangle edge length, x, is equal to the length of a face diagonal, as indicated in
2
Figure (a). And its length is related to the unit cell edge length,a, as
Excerpts from this work may be reproduced by instructorsdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has been aAny 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. 2 2 2 2
x = a + a = 2a
or
x = a 2
4R
For BCC, a = (Equation 3.3), and, therefore,
3
4R 2
x = 3
Also, from Figure (b), with respect to the length y we may write
2
y2 + ⎜x ⎟ = x 2
⎝2 ⎠
x 3
which leads toy = . And, substitution for the above expression for x yields
2
x 3 ⎛4R 2 ⎞⎛ 3⎞ 4R 2
y = = ⎜ ⎟⎜ ⎟ =
2 ⎝ 3 ⎠⎝ 2 ⎠ 2
Thus, the area of this triangle is equal to
⎛ ⎞⎛ ⎞ 2
1 ⎛1⎞ 4R 2 4R 2 8R
AREA = 2 xy = ⎝2⎠ ⎜ 3 ⎟⎜ 2 ⎟ = 3
⎝ ⎠⎝ ⎠
And, finally, the planar density for this (111) plane is
3
PD (BCC) = 0.5 atom = = 0.11
111 8R 2 16R 2 R2
3
Excerpts from this work may be reproduced by indistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook 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. 8.4 One slip system for the BCC crystal structure is { 110}111 . In a manner similar to Figure 8.6b,
sketch a {110}-type plane for the BCC structure, represen ting atom positions with circles. Now, using arrows,
indicate two different 111 slip directions within this plane.
Solution
Below is shown the atomic packing for a BCC {110} -type plane. The arrows indicate two different 111
type directions.
Excerpts from this work may be reproduced by instrudistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has beAny 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. Slip in Single Crystals
8.6 Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction
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 applied stress of 12 MPa (1750 psi) 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 = 60°, λ = 35°, and that the values of the
critical resolved shear stress and applied tensile stress ar e 6.2 MPa (900 psi) and 12 MPa (1750 psi), respectively.
From Equation 8.2
τR = σ cos φ cos λ = (12 MPa)(cos 60°)(cos 35°) = 4.91 MPa (717 psi)
Since the resolved shear stress (4.91 MPa) is less that the critical resolved shear stress (6.2 MPa), the single crystal
will not yield.
However, from Equation 8.4, the stress at which yielding occurs is
τ
σ = crss = 6.2 MPa = 15.1 MPa (2200 psi)
y cos φ cos λ (cos 60° )(cos 35°)
Excerpts from this work may be reproduced by instrudistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has beAny 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. 8.8 (a) A single crystal of a metal that has the BCC crys tal structure is oriented such that a tensile stress
is applied in the [100] direction. If the magnitude of this stress is 4.0 MPa, compute the resolved shear stress in the
[11 1] direction on each of the (110), (011), and(101 )planes.
(b) On the basis of these resolved shear stress va lues, which slip system(s) is (are) most favorably
oriented?
Solution
(a) This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the1 1]
direction on each of the (110), (011), and (101 ) planes. In order to solve this problem it is necessary to employ
Equation 8.2, which means that we first need to solve for the for angles λ and φ for the three slip systems.
For each of these thr ee slip systems, theλ will be the same—i.e., the angle between the direction of the
applied stress, [100] and the slip directi[11 1]. This angle λ may be determined using Equation 8.6
⎡ ⎤
−1⎢ u1 2+ v v1 2 + w1 2 ⎥
λ= cos ⎢ 2 2 2 2 2 2 ⎥
⎢ ()1 + v1 + w1 ()2 + v 2 + w 2 ⎥
⎣ ⎦
where (for [100]) u = 1 ,v = 0, w = 0, and (for [1 1 1) u = 1, v = –1, w = 1. Therefore, λ is determined as
1 1 1 2 2 2
⎡ ⎤
−1⎢ (1)(1)+ (0)(−1) + (0)(1) ⎥
λ= cos ⎢ 2 2 2 2 2 2 ⎥
⎢ []) + (0) + (0) [ ]) + (−1) + (1) ⎥
⎣ ⎦
−1 ⎛ 1 ⎞
= cos ⎜ ⎟= 54.7°
⎝ 3 ⎠
Let us now determine φ for the angle between the direction of the a pplied tensile stress—i.e., the [100] direction—
and the normal to the (110) slip plane—i.e., the [110] direction. Again, using Equation 8.6 where u = ,v = 0, w
1 1 1
= 0 (for [100]), and u = 1, v = 1, w = 0 (for [110]), φ is equal to
2 2 2
⎡ ⎤
−1⎢ (1)(1)+ (0)(1) + (0)(0) ⎥
φ[100]−[110] = cos ⎢ ⎥
⎢ []1) + (0) + (0) 2 []) + (1) + (0) 2 ⎥
⎣ ⎦
⎛ 1 ⎞
= cos−1 ⎜ ⎟= 45°
⎝ 2 ⎠
Excerpts from this work may be reproduced by instrdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has 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. Now, using Equation 8.2
τR = σ cosφ cosλ
we solve for the resolved shear stress for this slip system as
τ = (4.0 MPa) cos(45°)cos(54.7°) = (4.0 MPa)(0.707)(0.578) = 1.63 MPa
R(110)−[11 1] [ ]
Now, we must determine the value of φ for the (011)– [11 1]slip system—that is, the angle between the
direction of the applied stress, [100], and the normal to the (011) plane—i.e., the [011] direction. Again using
Equation 8.6
⎡ ⎤
−1⎢ (1)(0)+ (0)(1) + (0)(1) ⎥
λ [100]−[011] = cos ⎢ 2 2 2 2 2 2 ⎥
⎣ []1) + (0) + (0) []0) + (1) + (1) ⎦
= cos−1 (0) = 90°
Thus, the resolved shear stress for this (011[11 1] slip system is
τR (011)−[11]= (4.0 MPa) c[s(90°)cos(54.7°) = ]4.0 MPa)(0)(0.578) = 0 MPa
And, finally, it is necessary to determine the value of φ for the (101 )–[1 1 1]slip system —that is, the
angle between the direction of the app lied stress, [100], and the normal to the (101 )plane—i.e., the [101 ]
direction. Again using Equation 8.6
⎡ ⎤
−1 ⎢ (1)(1)+ (0)(0) + (0)(−1) ⎥
λ[100]−[10 1 ] cos ⎢ ⎥
⎢ []1) + (0) + (0) 2 [ ]) + (0) + (−1) 2 ⎥
⎣ ⎦
⎛ 1 ⎞
= cos −1⎜ ⎟ = 45°
⎝ 2⎠
Here, as with the (110)–[11 1] slip system above, the value of φ is 45°, which again leads to
Excerpts from this work may be reproduced by instrdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has bAny 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. τR = (4.0 MPa) c[ ]45°)cos(54.7°) = (4.0 MPa)(0.707)(0.578) = 1.63 MPa
(10 )−[11 1]
(b) The most favored slip system(s) is (a re) the one(s) that has (have) the largest τ value. Both (110)–
R
[1 1 1]and (101 )−[11 1] slip systems are most favored since they have the same τ (R.63 MPa), which is greater
than the τRvalue for (011)−[11 1] (viz., 0 MPa).
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students enrolled in courses for which the textbook has bAny 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. 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 in tension.
Solution
In order to determine the maximum possible yield strength for a single crystal of Cu pulled in tension, we
simply employ Equation 8.5 as
σ =2 τ = (2)(0.48 MPa) = 0.96 MPa (140 psi)
y crss
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Strengthening by Grain Size Reduction
8.10 Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip
process as are high-angle grain boundaries.
Answer
Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain
boundaries because there is not as much crystallographicmisalignment in the grain boundary region for small-angle,
and therefore not as much change in slip direction.
Excerpts from this work may be reproduced by instructordistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has been aAny 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
8.14 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing
their cross-sectional areas (while maintaining their circular cross ons). For one specimen, the initial and
deformed radii are 15 mm and 12 mm, respectively. The second 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.
Solution
In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed
to the same percent cold work. For the first specimen the percent cold workis computed using Equation 8.8 as
A − A π r2 −π r 2
%CW = 0 d × 100 = 0 d × 100
A0 πr0
2 2
π (15 mm) −π (12 mm)
= π (15 mm) 2 × 100 = 36%CW
For the second specimen, the deformed radius is computed using the aboveequation and solving for r
d
%CW
rd = r0 1 −
100
36%CW
= (11 mm) 1 − = 8.80 mm
100
Excerpts from this work may be reproduced by indistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook 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. Factors That Influence the Mechanical Properties of Semicrystalline
Polymers
Deformation of Elastomers
8.22 Briefly explain how each of the following influences the tensile or yield strength of a semicrystalline
polymer and why:
(a) Molecular weight
(b) Degree of crystallinity
(c) Deformation by drawing
(d) Annealing of an undeformed material
Solution
(a) The tensile strength of a semicrystalline polymer increases with increasing molecular weight. This
effect is explained by increased chain entanglements at higher molecular weights.
(b) Increasing the degree of crys tallinity of a semicrystalline polymer leads to an enhancement of the
tensile strength. Again, this is due to enhanced inte rchain bonding and forces; in response to applied stresses,
interchain motions are thus inhibited.
(c) Deformation by drawing increases the tensile stre ngth of a semicrystalline polymer. This effect is due
to the highly oriented chain structure th at is produced by drawing, which give s rise to higher interchain secondary
bonding forces.
(d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength.
Excerpts from this work may be reproduced by instructdistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has beenAny 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. 8.23 The tensile strength and number-average molecular weight for two poly(methyl methacrylate)
materials are as follows:
Tensile Strength Number-Average Molecular
(MPa) Weight (g/mol)
50 30,000
150 50,000
Estimate the tensile strength at a number-average molecular weight of 40,000 g/mol.
Solution
This problem gives us the tensile strengths aassociated number-average molecular weights for two
poly(methyl methacrylate) materials and then as ks that we estimate the tensile strMnn= 40,000 g/mol.
Equation 8.11 cites the dependen ce of the tensile strenMtn. Thus, using the data provided in the problem
statement, we may set up two simultaneous equations fro m which it is possible to solve for the two constants TS
∞
and A. These equations are as follows:
A
50 MPa = TS −∞ 30,000 g/mol
150 MPa = TS − A
∞ 50,000 g/mol
6
Thus, the values of the two constants aTS ∞ 300 MPa and A = 7.50 × 10 MPa-g/mol. Substituting these
values into Equation 8.11 fon = 40,000 g/mol leads to
A
TS = TS ∞
40,000 g/mol
7.50 × 10 MPa - g/mol
= 300 MPa −
40,000 g/mol
= 112.5 MPa
Excerpts from this work may be reproduced bydistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbooAny 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. 8.24 For each of the following pairs of polymers, do the fo llowing: (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 tensile modulus and then cite the reason(s) for y our choice; and (3) if it is not possible to decide, then
state why.
(a) Branched and atactic poly(vinyl chloride) with a we ight-average molecular weight of 100,000 g/mol;
linear and isotactic poly(vinyl chloride) having aweight-average molecular weight of 75,000 g/mol
(b) Random styrene-butadiene copolymer with 5% of po ssible sites crosslinked; block styrene-butadiene
copolymer with 10% of possible sites crosslinked
(c) Branched polyethylene with a number-average mo lecular weight of 100,000 g/mol; atactic
polypropylene with a number-average molecular weight of 150,000 g/mol
Solution
(a) Yes, it is possible. The linear and isotactic pol y(vinyl chloride) will display a greater tensile modulus.
Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures
will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's
crystallinity leads to an increase in its tensile modulus. In addition, te nsile modulus is independent of molecular
weight--the atactic/branched materialhas the higher molecular weight.
(b) Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will
have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers
of the same material. A higher degree of crystallinity favor s larger moduli. In addition, the block copolymer also
has a higher degree of crosslinking; increasing the amountof crosslinking also enhances the tensile modulus.
(c) No, it is not possible. Branched polyethylen e will tend to have a low de gree of crystallinity since
branched polymers don't normally crystallize. The atactic polypropylene proba bly also has a relatively low degree
of crystallinity; atactic structures al so don't tend to crystallize, and polypropy lene has a more complex repeat unit
structure than does polyethylene. Tens ile modulus increases with degree of crystallinity, and it is not possible to
determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight.
Excerpts from this work may be reproduced by instructordistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has been 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. 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 number-average molecular weight of 100,000 g/mol and 10% of available sites
crosslinked; polyisoprene having a number-average molecular weight of 100,000 g/mol and 20% of available sites
crosslinked
(b) Syndiotactic polypropylene having a weight-average molecular weight of 100,000 g/mol; atactic
polypropylene having a weight-average molecular weight of 75,000 g/mol
(c) Branched polyethylene having a num ber-average molecular weight of 90,000 g/mol; heavily
crosslinked polyethylene having a number-average molecular weight of 90,000 g/mol
Solution
(a) Shown below are the stress-strain curves for th e two polyisoprene materials, both of which have a
molecular weight of 100,000 g/mol. These two materials are elastomers and will have curves similar to curve C in
Figure 7.22. However, the curve for the material having the greater number of crossli nks (20%) will have a higher
elastic modulus at all strains.
(b) Shown below are the stress-strain curves for th e two polypropylene materials. These materials will
most probably display the stress-strain behavior of a normal plastic, curve B in Figure 7.22. However, the
syndiotactic polypropylene has a hi gher molecular weight and will also undoubtedly have a higher degree of
crystallinity; therefore, it will have a higher strength.
Excerpts from this work may be reproduced by instructodistribution on a not-for-profit ba sis for testing or instructiona l purposes only to
students enrolled in courses for which the textbook has beenAny 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. (c) Shown below are the stress-strain curves for the two polyethylene mate rials. The branched
polyethylene will display the behavior of a normal

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