Chap 7 Solns
CHAPTER 8
DEFORMATION AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS
Slip Systems 8.3 (a) Compare planar densities (Section 3.15 and Problem 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
PD110 (FCC) =
1 4 R2
2
0.177
=
R2
Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem W3.46, which are as follows: PD100 (FCC) =
PD111 (FCC) =
1 4 R2
1 2 R2
3
=
0.25
=
R2
0.29 R2
(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:
PD100 (BCC) =
PD110 (BCC) =
3 16R 2 3
8 R2
2
=
=
0.19 R2 0.27 R2
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Below is a BCC unit cell, within which is shown a (111) plane.
(a) The centers of the three corner atoms, denoted by A, B, and C lie on this 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 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, and 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 triangle edge length, x, is equal to the length of a face diagonal, as indicated in the area of this triangle is 2 Figure (a). And its length is related to the unit cell edge length, a, as 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.
x 2 = a 2 + a 2 = 2a 2
or x=a 2
For BCC, a =
4R (Equation 3.3), and, therefore, 3
x=
4R 2 3
Also, from Figure (b), with respect to the length y we may write ⎛ x ⎞2 y2 + ⎜ ⎟ = x2 ⎝2⎠
which leads to y =
x 3 . And, substitution for the above expression for x yields 2
y=
x 3 ⎛ 4 R 2 ⎞⎛ 3 ⎞ 4 R 2 = ⎜⎜ ⎟⎜ ⎟= 2 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 ⎝
Thus, the area of this triangle is equal to
AREA =
⎛ 1 ⎞ ⎛ 4 R 2 ⎞ ⎛ 4 R 2 ⎞ 8 R2 1 x y = ⎜ ⎟ ⎜⎜ ⎟⎜ ⎟= ⎝ 2 ⎠⎝ 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ 3 2
And, finally, the planar density for this (111) plane is
PD111 (BCC) =
0.5 atom 8 R2
=
3 16 R 2
=
0.11 R2
3
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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, representing 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.
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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 are 6.2 MPa (900 psi) and 12 MPa (1750 psi), respectively. From Equation 8.2
τ R = 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
σy =
τ crss
cos φ cos λ
=
6.2 MPa
(cos 60°)(cos 35°)
= 15.1 MPa (2200 psi)
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8.8 (a) A single crystal of a metal that has the BCC crystal 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 values, 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 the [11 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 three slip systems, the λ will be the same—i.e., the angle between the direction of the applied stress, [100] and the slip direction, [1 1 1] . This angle λ may be determined using Equation 8.6 ⎡ ⎢ λ = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ u12 + v12 + w12 u 22 + v 22 + w 22 ⎥ ⎦ u1u 2 + v1v 2 + w1w 2
(
)(
)
where (for [100]) u1 = 1, v1 = 0, w1 = 0, and (for [1 1 1] ) u2 = 1, v2 = –1, w2 = 1. Therefore, λ is determined as ⎡
⎢ λ = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (1) 2 + (−1) 2 + (1) 2 ⎥ ⎦ (1)(1) + (0)(−1) + (0)(1)
[
][
]
⎛ 1 ⎞ = cos−1 ⎜⎜ ⎟⎟ = 54.7° ⎝ 3⎠
Let us now determine φ for the angle between the direction of the applied 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 u1 = 1, v1 = 0, w1 = 0 (for [100]), and u2 = 1, v2 = 1, w2 = 0 (for [110]), φ is equal to ⎡
⎢ φ[100 ]−[110] = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ 2 2 2 2 2 2 (1) + (0) + (0) (1) + (1) + (0) ⎥ ⎦
[
(1)(1) + (0)(1) + (0)(0)
][
]
⎛ 1 ⎞ = cos−1 ⎜⎜ ⎟⎟ = 45° ⎝ 2⎠
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Now, using Equation 8.2
τ R = σ cosφ cos λ we solve for the resolved shear stress for this slip system as
τ R(110)−[1 1 1] = (4.0 MPa) [cos (45°) cos (54.7°)] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa Now, we must determine the value of φ for the (011)– [1 1 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 ⎡
⎢ λ[100 ]−[ 011] = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (0) 2 + (1) 2 + (1) 2 ⎥ ⎦
[
(1)(0) + (0)(1) + (0)(1)
][
]
= cos−1 (0) = 90°
Thus, the resolved shear stress for this (011)– [1 1 1] slip system is
τ R(011)−[1 1 1] = (4.0 MPa) [cos (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 ) – [11 1] slip system —that is, the angle between the direction of the applied stress, [100], and the normal to the (10 1 ) plane—i.e., the [10 1 ] direction. Again using Equation 8.6 ⎡ ⎢ λ = cos−1 ⎢ [100 ]−[101 ] ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (1) 2 + (0) 2 + (−1) 2 ⎥ ⎦
[
(1)(1) + (0)(0) + (0)(−1)
][
]
⎛ 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
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τ R(10 1 )−[1 1 1] = (4.0 MPa) [cos (45°) cos (54.7°)] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa (b) The most favored slip system(s) is (are) the one(s) that has (have) the largest τR value. Both (110)– [11 1] and (101) −[11 1] slip systems are most favored since they have the same τR (1.63 MPa), which is greater than the τR value for (011) −[11 1] (viz., 0 MPa).
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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 y = 2τ crss = (2)(0.48 MPa) = 0.96 MPa (140 psi)
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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 crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.
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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 sections). 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 work is computed using Equation 8.8 as
%CW =
=
A0 − Ad π r2 − π r2 × 100 = 0 2 d × 100 A0 π r0
π (15 mm)2 − π (12 mm)2 × 100 = 36%CW π (15 mm)2
For the second specimen, the deformed radius is computed using the above equation and solving for rd as rd = r0 1 −
= (11 mm) 1 −
%CW 100
36%CW = 8.80 mm 100
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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 crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Again, this is due to enhanced interchain bonding and forces; in response to applied stresses, interchain motions are thus inhibited. (c) Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary bonding forces. (d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength.
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8.23
The tensile strength and number-average molecular weight for two poly(methyl methacrylate)
materials are as follows:
Tensile Strength (MPa)
Number-Average Molecular 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 and associated number-average molecular weights for two poly(methyl methacrylate) materials and then asks that we estimate the tensile strength for M n = 40,000 g/mol. Equation 8.11 cites the dependence of the tensile strength on M n . Thus, using the data provided in the problem statement, we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ and A. These equations are as follows:
50 MPa = TS∞ −
A 30,000 g / mol
150 MPa = TS∞ −
A 50,000 g / mol
Thus, the values of the two constants are: TS∞ = 300 MPa and A = 7.50 × 106 MPa-g/mol. Substituting these values into Equation 8.11 for M n = 40,000 g/mol leads to
TS = TS∞ −
= 300 MPa −
A 40,000 g / mol
7.50 × 106 MPa - g / mol 40,000 g / mol
= 112.5 MPa
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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 tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol (b) Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene 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 number-average molecular weight of 150,000 g/mol Solution (a) Yes, it is possible. The linear and isotactic poly(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, tensile modulus is independent of molecular weight--the atactic/branched material has 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 favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus. (c) No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene. Tensile 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.
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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 number-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 the 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 crosslinks (20%) will have a higher elastic modulus at all strains.
(b) Shown below are the stress-strain curves for the 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 higher molecular weight and will also undoubtedly have a higher degree of crystallinity; therefore, it will have a higher strength.
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(c) Shown below are the stress-strain curves for the two polyethylene materials.
The branched
polyethylene will display the behavior of a normal plastic, curve B in Figure 7.22. On the other hand, the heavily crosslinked polyethylene will be stiffer, stronger, and more brittle (curve A of Figure 7.22).
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8.28 The vulcanization of polyisoprene is accomplished with sulfur atoms according to Equation 8.12. If 45.3 wt% sulfur is combined with polyisoprene, how many crosslinks will be associated with each isoprene repeat unit if it is assumed that, on the average, five sulfur atoms participate in each crosslink? Solution If we arbitrarily consider 100 g of the vulcanized material, 45.3 g will be sulfur and 54.7 g will be polyisoprene. Next, let us find how many moles of sulfur and isoprene correspond to these masses. The atomic weight of sulfur is 32.06 g/mol, and thus,
# moles S =
45.3 g = 1.41 mol 32.06 g / mol
Now, in each isoprene repeat unit there are five carbon atoms and eight hydrogen atoms. Thus, the molecular weight of a mole of isoprene units is (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol Or, in 54.7 g of polyisoprene, the number of moles is equal to
# moles isoprene =
54.7 g = 0.793 mol 68.11 g / mol
Therefore, the ratio of moles of S to the number of moles of polyisoprene is
1.41 mol : 1 = 1.78 : 1 0.793 mol When all possible sites are crosslinked, the ratio of the number of moles of sulfur to the number of moles of isoprene is 5:1; this is because there are two crosslink sites per repeat unit and each crosslink is shared between repeat units on adjacent chains, and there are 5 sulfur atoms per crosslink. Finally, to determine the fraction of sites that are crosslinked, we just divide the actual crosslinked sulfur/isoprene ratio by the completely crosslinked ratio. Or, fraction of repeat unit sites crosslinked =
1.78 /1 = 0.356 5/1
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8.29 Demonstrate, in a manner similar to Equation 8.12, how vulcanization may occur in a chloroprene rubber. Solution The reaction by which a chloroprene rubber may become vulcanized is as follows:
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DESIGN PROBLEMS 8.D3 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 deformation. 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. Solution First let us calculate the percent cold work and attendant tensile strength and ductility if the drawing is carried out without interruption. From Equation 8.8
%CW =
⎛ d ⎞2 ⎛ d ⎞2 π⎜ 0 ⎟ − π⎜ d ⎟ ⎝2⎠ ⎝2 ⎠ ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
× 100
⎛ 11.4 mm ⎞2 ⎛ 8.9 mm ⎞2 π⎜ ⎟ − π⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 × 100 = 40%CW = 2 ⎛ 11.4 mm ⎞ π⎜ ⎟ ⎝ ⎠ 2
At 40%CW, the steel will have a tensile strength on the order of 900 MPa (130,000 psi) [Figure 8.19(b)], which is adequate; however, the ductility will be less than 9%EL [Figure 8.19(c)], which is insufficient. Instead of performing the drawing in a single operation, let us initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold-work the material a second time in order to achieve the final diameter, tensile strength, and ductility. Reference to Figure 8.19(b) indicates that 17%CW is necessary to yield a tensile strength of 825 MPa (122,000 psi). Similarly, a maximum of 19%CW is possible for 12%EL [Figure 8.19(c)]. The average of these extremes is 18%CW. If the final diameter after the first drawing is d '0 , then
18%CW =
⎛ ' ⎞2 ⎛ 8.9 mm ⎞2 d π⎜ 0 ⎟ − π⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠ 2 ⎝ ⎠ ⎛ ' ⎞2 d π⎜ 0 ⎟ ⎜ 2 ⎟ ⎝ ⎠
× 100
And, solving for d 0' from this expression, yields
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d 0' =
8.9 mm 18%CW 1 − 100
= 9.83 mm (0.387 in.)
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CHAPTER 9
FAILURE
PROBLEM SOLUTIONS
Principles of Fracture Mechanics 9.1 What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10–4 mm (7.5 × 10–6 in.) and a crack length of 3.8 × 10–2 mm (1.5 × 10–3 in .) when a tensile stress of 140 MPa (20,000 psi) is applied? Solution This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation 9.1 is employed to solve this problem, as ⎛ a ⎞1/ 2 σ m = 2σ 0 ⎜⎜ ⎟⎟ ⎝ ρt ⎠
⎡ 3.8 × 10−2 mm ⎤1/2 ⎢ ⎥ 2 ⎥ = 2800 MPa (400,000 psi) = (2)(140 MPa) ⎢ ⎢ 1.9 × 10−4 mm ⎥ ⎢⎣ ⎥⎦
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9.3 An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of MgO is 1.0 J/m2. Data found in Table 7.1 may prove helpful.
Solution The maximum allowable surface crack length for MgO may be determined using Equation 9.3; taking 225 GPa as the modulus of elasticity (Table 7.1), and solving for a, leads to
a=
2 E γs
= π σ 2c
(2) (225 × 109 N / m2 ) (1.0 N / m) (π) (13.5 × 106 N / m2 )
2
= 7.9 × 10-4 m = 0.79 mm (0.031 in.)
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9.4 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa m (36.4 ksi in. ). It has been determined that fracture results at a stress of 300 MPa (43,500 psi) when the maximum (or critical) internal crack length is 4.0 mm (0.16 in.). For this same component and alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0 mm (0.24 in.)? Why or why not?
Solution It first becomes necessary to solve for the parameter Y, using Equation 9.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and a = 4.0 mm). Therefore,
Y =
K Ic = σ πa
40 MPa m = 1.68 ⎛ 4 × 10−3 m ⎞ (300 MPa) (π) ⎜ ⎟ 2 ⎝ ⎠
Now we will solve for the product Y σ πa for the other set of conditions, so as to ascertain whether or not this value is greater than the KIc for the alloy. Thus, ⎛ 6 × 10−3 m ⎞ Y σ π a = (1.68)(260 MPa) (π) ⎜ ⎟ 2 ⎠ ⎝
= 42.4 MPa m Therefore, fracture will occur since this value (42.4 MPa
(39 ksi in.) m ) is greater than the KIc of the material, 40 MPa m .
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9.5 A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m (75.0 ksi in. ). If, during service use, the plate is exposed to a tensile stress of 345 MPa (50,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.
Solution For this problem, we are given values of KIc (82.4 MPa
m ) , σ (345 MPa), and Y (1.0) for a large plate
and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for ac using Equation 9.7; therefore 2 2 1 ⎛ K Ic ⎞ 1 ⎡ 82.4 MPa m ⎤ ac = ⎜ ⎥ = 0.0182 m = 18.2 mm (0.72 in.) ⎟ = ⎢ π ⎝Y σ ⎠ π ⎣ (1.0)(345 MPa) ⎦
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9.6 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa m (90 ksi in. ) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Solution This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m ) , the design stress (σy/2 in which σy = 860 MPa), and Y = 1.0. We first need to compute the value of ac using Equation 9.7; thus ⎡ ⎤2 ⎢ ⎥ 1 ⎛K ⎞ 1 98.9 MPa m ⎥ ac = ⎜ Ic ⎟ = ⎢ = 0.0168 m = 16.8 mm (0.66 in.) ⎛ 860 MPa ⎞ ⎥ π ⎝ Yσ ⎠ π⎢ (1.0) ⎜ ⎟⎥ ⎢⎣ ⎝ ⎠⎦ 2 2
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
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Fracture of Ceramics Fracture of Polymers 9.8
The tensile strength of brittle materials may be determined using a variation of Equation 9.1.
Compute the critical crack tip radius for a glass specimen that experiences tensile fracture at an applied stress of 70 MPa (10,000 psi). Assume a critical surface crack length of 10–2 mm and a theoretical fracture strength of E/10, where E is the modulus of elasticity.
Solution We are asked for the critical crack tip radius for a glass. From Equation 9.1
σm
⎛ a ⎞1/2 = 2 σ 0 ⎜⎜ ⎟⎟ ⎝ ρt ⎠
Fracture will occur when σm reaches the fracture strength of the material, which is given as E/10; thus
⎛ a ⎞1/2 E = 2 σ 0 ⎜⎜ ⎟⎟ 10 ⎝ ρt ⎠ Or, solving for ρt
ρt =
400 aσ 20 E2
From Table 7.1, E = 69 GPa, and thus, ρt =
(400 ) (1 × 10−2 mm) (70 MPa)2
(69
× 103 MPa ) 2
= 4.1 × 10-6 mm = 4.1 nm
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9.11 A 6.4 mm (0.25 in.) diameter cylindrical rod fabricated from a 2014-T6 aluminum alloy (Figure 9.41) is subjected to reversed tension-compression load cycling along its axis.
If the maximum tensile and
compressive loads are +5340 N (+1200 lbf) and –5340 N (–1200 lbf) , respectively, determine its fatigue life. Assume that the stress plotted in Figure 9.41 is stress amplitude. Solution We are asked to determine the fatigue life for a cylindrical 2014-T6 aluminum rod given its diameter (6.4 mm) and the maximum tensile and compressive loads (+5340 N and –5340 N, respectively). The first thing that is necessary is to calculate values of σmax and σmin using Equation 7.1. Thus
σ max =
=
5340 N ⎛ 6.4 × 10−3 m ⎞2 (π) ⎜ ⎟ 2 ⎝ ⎠
Fmax Fmax = A0 ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
= 166 × 106 N/m2 = 166 MPa (24, 400 psi)
σ min =
=
−5340 N ⎛ 6.4 × 10−3 m ⎞2 (π) ⎜ ⎟ 2 ⎝ ⎠
Fmin ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
= − 166 × 106 N/m2 = − 166 MPa (−24, 400 psi)
Now it becomes necessary to compute the stress amplitude using Equation 9.17 as
σa =
σ max − σ min 2
=
166 MPa − (−166 MPa) = 166 MPa (24, 400 psi) 2
From Figure 9.41, for the 2014-T6 aluminum, the number of cycles to failure at this stress amplitude is about 1 × 107 cycles.
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9.12 The fatigue data for a steel alloy are given as follows:
Stress Amplitude [MPa (ksi)]
Cycles to Failure
470 (68.0)
104
440 (63.4)
3 × 104
390 (56.2)
105
350 (51.0)
3 × 105
310 (45.3)
106
290 (42.2)
3 × 106
290 (42.2)
107
290 (42.2)
108
(a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 415 MPa (60,000 psi) and 275 MPa (40,000 psi). (d) Estimate fatigue strengths at 2 × 104 and 6 × 105 cycles. Solution (a) The fatigue data for this alloy are plotted below.
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(b) The fatigue limit is the stress level at which the curve becomes horizontal, which is 290 MPa (42,200 psi). (c) From the plot, the fatigue lifetimes at a stress amplitude of 415 MPa (60,000 psi) is about 50,000 cycles (log N = 4.7). At 275 MPa (40,000 psi) the fatigue lifetime is essentially an infinite number of cycles since this stress amplitude is below the fatigue limit. (d) Also from the plot, the fatigue strengths at 2 × 104 cycles (log N = 4.30) and 6 × 105 cycles (log N = 5.78) are 440 MPa (64,000 psi) and 325 MPa (47,500 psi), respectively.
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9.15 (a) Compare the fatigue limits for PMMA (Figure 9.27) and the steel alloy for which fatigue data are given in Problem 9.12. (b) Compare the fatigue strengths at 106 cycles for nylon 6 (Figure 9.27) and 2014-T6 aluminum (Figure 9.41). Solution (a) The fatigue limits for PMMA and the steel alloy are 10 MPa (1450 psi) and 290 MPa (42,200 psi), respectively. (b) At 106 cycles, the fatigue strengths for nylon 6 and 2014-T6 aluminum are 11 MPa (1600 psi) and 200 MPa (30,000 psi ), respectively.
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Factors That Affect Fatigue Life 9.16 List four measures that may be taken to increase the resistance to fatigue of a metal alloy. Solution Four measures that may be taken to increase the fatigue resistance of a metal alloy are: (1) Polish the surface to remove stress amplification sites. (2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques. (3) Modify the design to eliminate notches and sudden contour changes. (4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening.
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Generalized Creep Behavior 9.17 The following creep data were taken on an aluminum alloy at 480°C (900°F) and a constant stress of 2.75 MPa (400 psi). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included.
Time (min)
Strain
Time (min)
Strain
0
0.00
18
0.82
2
0.22
20
0.88
4
0.34
22
0.95
6
0.41
24
1.03
8
0.48
26
1.12
10
0.55
28
1.22
12
0.62
30
1.36
14
0.68
32
1.53
16
0.75
34
1.77
Solution These creep data are plotted below
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The steady-state creep rate (Δε/Δt) is the slope of the linear region (i.e., the straight line that has been superimposed on the curve) as 1.20 − 0.25 Δε = = 3.2 × 10-2 min-1 30 min − 0 min Δt
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DEFORMATION AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS
Slip Systems 8.3 (a) Compare planar densities (Section 3.15 and Problem 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
PD110 (FCC) =
1 4 R2
2
0.177
=
R2
Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem W3.46, which are as follows: PD100 (FCC) =
PD111 (FCC) =
1 4 R2
1 2 R2
3
=
0.25
=
R2
0.29 R2
(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:
PD100 (BCC) =
PD110 (BCC) =
3 16R 2 3
8 R2
2
=
=
0.19 R2 0.27 R2
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Below is a BCC unit cell, within which is shown a (111) plane.
(a) The centers of the three corner atoms, denoted by A, B, and C lie on this 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 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, and 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 triangle edge length, x, is equal to the length of a face diagonal, as indicated in the area of this triangle is 2 Figure (a). And its length is related to the unit cell edge length, a, as 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.
x 2 = a 2 + a 2 = 2a 2
or x=a 2
For BCC, a =
4R (Equation 3.3), and, therefore, 3
x=
4R 2 3
Also, from Figure (b), with respect to the length y we may write ⎛ x ⎞2 y2 + ⎜ ⎟ = x2 ⎝2⎠
which leads to y =
x 3 . And, substitution for the above expression for x yields 2
y=
x 3 ⎛ 4 R 2 ⎞⎛ 3 ⎞ 4 R 2 = ⎜⎜ ⎟⎜ ⎟= 2 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 ⎝
Thus, the area of this triangle is equal to
AREA =
⎛ 1 ⎞ ⎛ 4 R 2 ⎞ ⎛ 4 R 2 ⎞ 8 R2 1 x y = ⎜ ⎟ ⎜⎜ ⎟⎜ ⎟= ⎝ 2 ⎠⎝ 3 ⎟⎠ ⎜⎝ 2 ⎟⎠ 3 2
And, finally, the planar density for this (111) plane is
PD111 (BCC) =
0.5 atom 8 R2
=
3 16 R 2
=
0.11 R2
3
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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, representing 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.
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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 are 6.2 MPa (900 psi) and 12 MPa (1750 psi), respectively. From Equation 8.2
τ R = 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
σy =
τ crss
cos φ cos λ
=
6.2 MPa
(cos 60°)(cos 35°)
= 15.1 MPa (2200 psi)
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8.8 (a) A single crystal of a metal that has the BCC crystal 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 values, 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 the [11 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 three slip systems, the λ will be the same—i.e., the angle between the direction of the applied stress, [100] and the slip direction, [1 1 1] . This angle λ may be determined using Equation 8.6 ⎡ ⎢ λ = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ u12 + v12 + w12 u 22 + v 22 + w 22 ⎥ ⎦ u1u 2 + v1v 2 + w1w 2
(
)(
)
where (for [100]) u1 = 1, v1 = 0, w1 = 0, and (for [1 1 1] ) u2 = 1, v2 = –1, w2 = 1. Therefore, λ is determined as ⎡
⎢ λ = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (1) 2 + (−1) 2 + (1) 2 ⎥ ⎦ (1)(1) + (0)(−1) + (0)(1)
[
][
]
⎛ 1 ⎞ = cos−1 ⎜⎜ ⎟⎟ = 54.7° ⎝ 3⎠
Let us now determine φ for the angle between the direction of the applied 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 u1 = 1, v1 = 0, w1 = 0 (for [100]), and u2 = 1, v2 = 1, w2 = 0 (for [110]), φ is equal to ⎡
⎢ φ[100 ]−[110] = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ 2 2 2 2 2 2 (1) + (0) + (0) (1) + (1) + (0) ⎥ ⎦
[
(1)(1) + (0)(1) + (0)(0)
][
]
⎛ 1 ⎞ = cos−1 ⎜⎜ ⎟⎟ = 45° ⎝ 2⎠
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Now, using Equation 8.2
τ R = σ cosφ cos λ we solve for the resolved shear stress for this slip system as
τ R(110)−[1 1 1] = (4.0 MPa) [cos (45°) cos (54.7°)] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa Now, we must determine the value of φ for the (011)– [1 1 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 ⎡
⎢ λ[100 ]−[ 011] = cos−1 ⎢ ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (0) 2 + (1) 2 + (1) 2 ⎥ ⎦
[
(1)(0) + (0)(1) + (0)(1)
][
]
= cos−1 (0) = 90°
Thus, the resolved shear stress for this (011)– [1 1 1] slip system is
τ R(011)−[1 1 1] = (4.0 MPa) [cos (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 ) – [11 1] slip system —that is, the angle between the direction of the applied stress, [100], and the normal to the (10 1 ) plane—i.e., the [10 1 ] direction. Again using Equation 8.6 ⎡ ⎢ λ = cos−1 ⎢ [100 ]−[101 ] ⎢ ⎣
⎤ ⎥ ⎥ (1) 2 + (0) 2 + (0) 2 (1) 2 + (0) 2 + (−1) 2 ⎥ ⎦
[
(1)(1) + (0)(0) + (0)(−1)
][
]
⎛ 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
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τ R(10 1 )−[1 1 1] = (4.0 MPa) [cos (45°) cos (54.7°)] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa (b) The most favored slip system(s) is (are) the one(s) that has (have) the largest τR value. Both (110)– [11 1] and (101) −[11 1] slip systems are most favored since they have the same τR (1.63 MPa), which is greater than the τR value for (011) −[11 1] (viz., 0 MPa).
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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 y = 2τ crss = (2)(0.48 MPa) = 0.96 MPa (140 psi)
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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 crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.
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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 sections). 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 work is computed using Equation 8.8 as
%CW =
=
A0 − Ad π r2 − π r2 × 100 = 0 2 d × 100 A0 π r0
π (15 mm)2 − π (12 mm)2 × 100 = 36%CW π (15 mm)2
For the second specimen, the deformed radius is computed using the above equation and solving for rd as rd = r0 1 −
= (11 mm) 1 −
%CW 100
36%CW = 8.80 mm 100
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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 crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Again, this is due to enhanced interchain bonding and forces; in response to applied stresses, interchain motions are thus inhibited. (c) Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary bonding forces. (d) Annealing an undeformed semicrystalline polymer produces an increase in its tensile strength.
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8.23
The tensile strength and number-average molecular weight for two poly(methyl methacrylate)
materials are as follows:
Tensile Strength (MPa)
Number-Average Molecular 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 and associated number-average molecular weights for two poly(methyl methacrylate) materials and then asks that we estimate the tensile strength for M n = 40,000 g/mol. Equation 8.11 cites the dependence of the tensile strength on M n . Thus, using the data provided in the problem statement, we may set up two simultaneous equations from which it is possible to solve for the two constants TS∞ and A. These equations are as follows:
50 MPa = TS∞ −
A 30,000 g / mol
150 MPa = TS∞ −
A 50,000 g / mol
Thus, the values of the two constants are: TS∞ = 300 MPa and A = 7.50 × 106 MPa-g/mol. Substituting these values into Equation 8.11 for M n = 40,000 g/mol leads to
TS = TS∞ −
= 300 MPa −
A 40,000 g / mol
7.50 × 106 MPa - g / mol 40,000 g / mol
= 112.5 MPa
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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 tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol (b) Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene 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 number-average molecular weight of 150,000 g/mol Solution (a) Yes, it is possible. The linear and isotactic poly(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, tensile modulus is independent of molecular weight--the atactic/branched material has 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 favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus. (c) No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene. Tensile 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.
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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 number-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 the 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 crosslinks (20%) will have a higher elastic modulus at all strains.
(b) Shown below are the stress-strain curves for the 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 higher molecular weight and will also undoubtedly have a higher degree of crystallinity; therefore, it will have a higher strength.
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(c) Shown below are the stress-strain curves for the two polyethylene materials.
The branched
polyethylene will display the behavior of a normal plastic, curve B in Figure 7.22. On the other hand, the heavily crosslinked polyethylene will be stiffer, stronger, and more brittle (curve A of Figure 7.22).
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8.28 The vulcanization of polyisoprene is accomplished with sulfur atoms according to Equation 8.12. If 45.3 wt% sulfur is combined with polyisoprene, how many crosslinks will be associated with each isoprene repeat unit if it is assumed that, on the average, five sulfur atoms participate in each crosslink? Solution If we arbitrarily consider 100 g of the vulcanized material, 45.3 g will be sulfur and 54.7 g will be polyisoprene. Next, let us find how many moles of sulfur and isoprene correspond to these masses. The atomic weight of sulfur is 32.06 g/mol, and thus,
# moles S =
45.3 g = 1.41 mol 32.06 g / mol
Now, in each isoprene repeat unit there are five carbon atoms and eight hydrogen atoms. Thus, the molecular weight of a mole of isoprene units is (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol Or, in 54.7 g of polyisoprene, the number of moles is equal to
# moles isoprene =
54.7 g = 0.793 mol 68.11 g / mol
Therefore, the ratio of moles of S to the number of moles of polyisoprene is
1.41 mol : 1 = 1.78 : 1 0.793 mol When all possible sites are crosslinked, the ratio of the number of moles of sulfur to the number of moles of isoprene is 5:1; this is because there are two crosslink sites per repeat unit and each crosslink is shared between repeat units on adjacent chains, and there are 5 sulfur atoms per crosslink. Finally, to determine the fraction of sites that are crosslinked, we just divide the actual crosslinked sulfur/isoprene ratio by the completely crosslinked ratio. Or, fraction of repeat unit sites crosslinked =
1.78 /1 = 0.356 5/1
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8.29 Demonstrate, in a manner similar to Equation 8.12, how vulcanization may occur in a chloroprene rubber. Solution The reaction by which a chloroprene rubber may become vulcanized is as follows:
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DESIGN PROBLEMS 8.D3 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 deformation. 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. Solution First let us calculate the percent cold work and attendant tensile strength and ductility if the drawing is carried out without interruption. From Equation 8.8
%CW =
⎛ d ⎞2 ⎛ d ⎞2 π⎜ 0 ⎟ − π⎜ d ⎟ ⎝2⎠ ⎝2 ⎠ ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
× 100
⎛ 11.4 mm ⎞2 ⎛ 8.9 mm ⎞2 π⎜ ⎟ − π⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 × 100 = 40%CW = 2 ⎛ 11.4 mm ⎞ π⎜ ⎟ ⎝ ⎠ 2
At 40%CW, the steel will have a tensile strength on the order of 900 MPa (130,000 psi) [Figure 8.19(b)], which is adequate; however, the ductility will be less than 9%EL [Figure 8.19(c)], which is insufficient. Instead of performing the drawing in a single operation, let us initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold-work the material a second time in order to achieve the final diameter, tensile strength, and ductility. Reference to Figure 8.19(b) indicates that 17%CW is necessary to yield a tensile strength of 825 MPa (122,000 psi). Similarly, a maximum of 19%CW is possible for 12%EL [Figure 8.19(c)]. The average of these extremes is 18%CW. If the final diameter after the first drawing is d '0 , then
18%CW =
⎛ ' ⎞2 ⎛ 8.9 mm ⎞2 d π⎜ 0 ⎟ − π⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠ 2 ⎝ ⎠ ⎛ ' ⎞2 d π⎜ 0 ⎟ ⎜ 2 ⎟ ⎝ ⎠
× 100
And, solving for d 0' from this expression, yields
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d 0' =
8.9 mm 18%CW 1 − 100
= 9.83 mm (0.387 in.)
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CHAPTER 9
FAILURE
PROBLEM SOLUTIONS
Principles of Fracture Mechanics 9.1 What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10–4 mm (7.5 × 10–6 in.) and a crack length of 3.8 × 10–2 mm (1.5 × 10–3 in .) when a tensile stress of 140 MPa (20,000 psi) is applied? Solution This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation 9.1 is employed to solve this problem, as ⎛ a ⎞1/ 2 σ m = 2σ 0 ⎜⎜ ⎟⎟ ⎝ ρt ⎠
⎡ 3.8 × 10−2 mm ⎤1/2 ⎢ ⎥ 2 ⎥ = 2800 MPa (400,000 psi) = (2)(140 MPa) ⎢ ⎢ 1.9 × 10−4 mm ⎥ ⎢⎣ ⎥⎦
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9.3 An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of MgO is 1.0 J/m2. Data found in Table 7.1 may prove helpful.
Solution The maximum allowable surface crack length for MgO may be determined using Equation 9.3; taking 225 GPa as the modulus of elasticity (Table 7.1), and solving for a, leads to
a=
2 E γs
= π σ 2c
(2) (225 × 109 N / m2 ) (1.0 N / m) (π) (13.5 × 106 N / m2 )
2
= 7.9 × 10-4 m = 0.79 mm (0.031 in.)
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9.4 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa m (36.4 ksi in. ). It has been determined that fracture results at a stress of 300 MPa (43,500 psi) when the maximum (or critical) internal crack length is 4.0 mm (0.16 in.). For this same component and alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0 mm (0.24 in.)? Why or why not?
Solution It first becomes necessary to solve for the parameter Y, using Equation 9.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and a = 4.0 mm). Therefore,
Y =
K Ic = σ πa
40 MPa m = 1.68 ⎛ 4 × 10−3 m ⎞ (300 MPa) (π) ⎜ ⎟ 2 ⎝ ⎠
Now we will solve for the product Y σ πa for the other set of conditions, so as to ascertain whether or not this value is greater than the KIc for the alloy. Thus, ⎛ 6 × 10−3 m ⎞ Y σ π a = (1.68)(260 MPa) (π) ⎜ ⎟ 2 ⎠ ⎝
= 42.4 MPa m Therefore, fracture will occur since this value (42.4 MPa
(39 ksi in.) m ) is greater than the KIc of the material, 40 MPa m .
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9.5 A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m (75.0 ksi in. ). If, during service use, the plate is exposed to a tensile stress of 345 MPa (50,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.
Solution For this problem, we are given values of KIc (82.4 MPa
m ) , σ (345 MPa), and Y (1.0) for a large plate
and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for ac using Equation 9.7; therefore 2 2 1 ⎛ K Ic ⎞ 1 ⎡ 82.4 MPa m ⎤ ac = ⎜ ⎥ = 0.0182 m = 18.2 mm (0.72 in.) ⎟ = ⎢ π ⎝Y σ ⎠ π ⎣ (1.0)(345 MPa) ⎦
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9.6 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa m (90 ksi in. ) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Solution This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m ) , the design stress (σy/2 in which σy = 860 MPa), and Y = 1.0. We first need to compute the value of ac using Equation 9.7; thus ⎡ ⎤2 ⎢ ⎥ 1 ⎛K ⎞ 1 98.9 MPa m ⎥ ac = ⎜ Ic ⎟ = ⎢ = 0.0168 m = 16.8 mm (0.66 in.) ⎛ 860 MPa ⎞ ⎥ π ⎝ Yσ ⎠ π⎢ (1.0) ⎜ ⎟⎥ ⎢⎣ ⎝ ⎠⎦ 2 2
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
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Fracture of Ceramics Fracture of Polymers 9.8
The tensile strength of brittle materials may be determined using a variation of Equation 9.1.
Compute the critical crack tip radius for a glass specimen that experiences tensile fracture at an applied stress of 70 MPa (10,000 psi). Assume a critical surface crack length of 10–2 mm and a theoretical fracture strength of E/10, where E is the modulus of elasticity.
Solution We are asked for the critical crack tip radius for a glass. From Equation 9.1
σm
⎛ a ⎞1/2 = 2 σ 0 ⎜⎜ ⎟⎟ ⎝ ρt ⎠
Fracture will occur when σm reaches the fracture strength of the material, which is given as E/10; thus
⎛ a ⎞1/2 E = 2 σ 0 ⎜⎜ ⎟⎟ 10 ⎝ ρt ⎠ Or, solving for ρt
ρt =
400 aσ 20 E2
From Table 7.1, E = 69 GPa, and thus, ρt =
(400 ) (1 × 10−2 mm) (70 MPa)2
(69
× 103 MPa ) 2
= 4.1 × 10-6 mm = 4.1 nm
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9.11 A 6.4 mm (0.25 in.) diameter cylindrical rod fabricated from a 2014-T6 aluminum alloy (Figure 9.41) is subjected to reversed tension-compression load cycling along its axis.
If the maximum tensile and
compressive loads are +5340 N (+1200 lbf) and –5340 N (–1200 lbf) , respectively, determine its fatigue life. Assume that the stress plotted in Figure 9.41 is stress amplitude. Solution We are asked to determine the fatigue life for a cylindrical 2014-T6 aluminum rod given its diameter (6.4 mm) and the maximum tensile and compressive loads (+5340 N and –5340 N, respectively). The first thing that is necessary is to calculate values of σmax and σmin using Equation 7.1. Thus
σ max =
=
5340 N ⎛ 6.4 × 10−3 m ⎞2 (π) ⎜ ⎟ 2 ⎝ ⎠
Fmax Fmax = A0 ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
= 166 × 106 N/m2 = 166 MPa (24, 400 psi)
σ min =
=
−5340 N ⎛ 6.4 × 10−3 m ⎞2 (π) ⎜ ⎟ 2 ⎝ ⎠
Fmin ⎛ d ⎞2 π⎜ 0 ⎟ ⎝2⎠
= − 166 × 106 N/m2 = − 166 MPa (−24, 400 psi)
Now it becomes necessary to compute the stress amplitude using Equation 9.17 as
σa =
σ max − σ min 2
=
166 MPa − (−166 MPa) = 166 MPa (24, 400 psi) 2
From Figure 9.41, for the 2014-T6 aluminum, the number of cycles to failure at this stress amplitude is about 1 × 107 cycles.
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9.12 The fatigue data for a steel alloy are given as follows:
Stress Amplitude [MPa (ksi)]
Cycles to Failure
470 (68.0)
104
440 (63.4)
3 × 104
390 (56.2)
105
350 (51.0)
3 × 105
310 (45.3)
106
290 (42.2)
3 × 106
290 (42.2)
107
290 (42.2)
108
(a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 415 MPa (60,000 psi) and 275 MPa (40,000 psi). (d) Estimate fatigue strengths at 2 × 104 and 6 × 105 cycles. Solution (a) The fatigue data for this alloy are plotted below.
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(b) The fatigue limit is the stress level at which the curve becomes horizontal, which is 290 MPa (42,200 psi). (c) From the plot, the fatigue lifetimes at a stress amplitude of 415 MPa (60,000 psi) is about 50,000 cycles (log N = 4.7). At 275 MPa (40,000 psi) the fatigue lifetime is essentially an infinite number of cycles since this stress amplitude is below the fatigue limit. (d) Also from the plot, the fatigue strengths at 2 × 104 cycles (log N = 4.30) and 6 × 105 cycles (log N = 5.78) are 440 MPa (64,000 psi) and 325 MPa (47,500 psi), respectively.
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9.15 (a) Compare the fatigue limits for PMMA (Figure 9.27) and the steel alloy for which fatigue data are given in Problem 9.12. (b) Compare the fatigue strengths at 106 cycles for nylon 6 (Figure 9.27) and 2014-T6 aluminum (Figure 9.41). Solution (a) The fatigue limits for PMMA and the steel alloy are 10 MPa (1450 psi) and 290 MPa (42,200 psi), respectively. (b) At 106 cycles, the fatigue strengths for nylon 6 and 2014-T6 aluminum are 11 MPa (1600 psi) and 200 MPa (30,000 psi ), respectively.
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Factors That Affect Fatigue Life 9.16 List four measures that may be taken to increase the resistance to fatigue of a metal alloy. Solution Four measures that may be taken to increase the fatigue resistance of a metal alloy are: (1) Polish the surface to remove stress amplification sites. (2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques. (3) Modify the design to eliminate notches and sudden contour changes. (4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening.
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Generalized Creep Behavior 9.17 The following creep data were taken on an aluminum alloy at 480°C (900°F) and a constant stress of 2.75 MPa (400 psi). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included.
Time (min)
Strain
Time (min)
Strain
0
0.00
18
0.82
2
0.22
20
0.88
4
0.34
22
0.95
6
0.41
24
1.03
8
0.48
26
1.12
10
0.55
28
1.22
12
0.62
30
1.36
14
0.68
32
1.53
16
0.75
34
1.77
Solution These creep data are plotted below
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The steady-state creep rate (Δε/Δt) is the slope of the linear region (i.e., the straight line that has been superimposed on the curve) as 1.20 − 0.25 Δε = = 3.2 × 10-2 min-1 30 min − 0 min Δt
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