Download Edge Dislocations - Materials Science Engineering - Solved Quiz and more Exercises Material Engineering in PDF only on Docsity!
7.2 When the two edge dislocations become aligned, a planar region of vacancies will exist between the dislocations as:
7.3 It is possible for two screw dislocations of opposite sign to annihilate one another if their dislocation lines are parallel. This is demonstrated in the figure below.
7.6 (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation (3.12) as
PD 110 (FCC) = 1 4 R^2
R^2
Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.47, which are as follows:
PD 100 (FCC) = 1 4R 2
R^2
PD 111 (FCC) = 1
2R 2 3
R^2
(b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.48, which are as follows:
PD 100 (BCC) = 3 16 R^2
R 2
PD 110 (BCC) = 3
8R 2 2
R 2
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)
PD 111 (BCC) = 0.5 atom 8R 2 3
16R^2
R^2
7.9 Resolved shear stress is the shear component of an applied tensile (or compressive) stress resolved along a slip plane that is other than perpendicular or parallel to the stress axis. The critical resolved shear stress is the value of resolved shear stress at which yielding begins; it is a property of the material.
7.12 We are asked to compute the critical resolved shear stress for Zn. As stipulated in the problem, φ = 65°, while possible values for λ are 30°, 48°, and 78°. (a) Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. The cosines for the possible λ values are given below.
cos(30°) = 0. cos(48°) = 0. cos(78°) = 0.
Thus, the slip direction is at an angle of 30° with the tensile axis. (b) From Equation (7.3), the critical resolved shear stress is just
τcrss = σy (cos φ cos λ)max
= (2.5 MPa) cos(65[ °)cos( 3 0°)] = 0.90 MPa (130 psi)
7.20 These three strengthening mechanisms are described in Sections 7.8, 7.9, and 7.10.
7.21 (a) Perhaps the easiest way to solve for σ o and ky in Equation (7.5) is to pick two values
each of σ y and d -^ 1/2^ from Figure 7.13, and then set up and solve two simultaneous equations. For example
d-^ 1/2^ (mm)-^ 1/2^ σy (MPa) 4 75 12 175
The two equations are thus
75 = σo + 4k (^) y
175 = σo + 12k (^) y
These yield the values of
k y = 12.5 MPa (mm)1/2^ [1810 psi mm ( )1/ 2]
σo = 25 MPa (3630 psi)
(b) When d = 1.0 x 10-3 mm, d -^ 1/2^ = 31.6 mm-^ 1/2, and, using Equation (7.5),
σy = σo + k (^) y d -1/
= (25 MPa) + 12.5 [ MPa mm( )1/ 2] (31.6 mm -1/2)= 420 MPa (61,000 psi)
7.22 We are asked to determine the grain diameter for an iron which will give a yield strength of 205 MPa (30,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of Equation (7.5), solve for σ o and ky , and finally determine the value of d when σ y = 205 MPa. The data pertaining to this problem may be tabulated as follows:
σy d (mm) d-^ 1/2^ (mm)-^ 1/ 135 MPa 5 x 10-2^ 4. 260 MPa 8 x 10-3^ 11.
The two equations thus become
135 MPa = σo + (4.47) k (^) y 260 MPa = σo + (11.18) k (^) y
Which yield the values, σ o = 51.7 MPa and ky = 18.63 MPa(mm) 1/2. At a yield strength of 205 MPa
7.31 We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 10^6 mm-2. It is first necessary to compute the value of the constant A from the one set of data as
A =
τcrss − τo ρD^ =^
0.69 MPa − 0.069 MPa 10 4 mm−^2
= 6.21 x 10−^3 MPa − mm (0.90 psi − mm)
Now, the critical resolved shear stress may be determined at a dislocation density of 10^6 mm-2 as
τcrss = τo + A ρD
= (0.069 MPa) + (^) (6.21 x 10 -3^ MPa - mm) 106 mm−^2 = 6.28 MPa (910 psi)
7.32 For recovery, there is some relief of internal strain energy by dislocation motion; however, there are virtually no changes in either the grain structure or mechanical characteristics. During recrystallization, on the other hand, a new set of strain-free grains forms, and the material becomes softer and more ductile.
7.34 During cold-working, the grain structure of the metal has been distorted to accommodate the deformation. Recrystallization produces grains that are equiaxed and smaller than the parent grains.
7.35 Metals such as lead and tin do not strain harden at room temperature because their recrystallization temperatures lie below room temperature (Table 7.2).
7.36 (a) The driving force for recrystallization is the difference in internal energy between the strained and unstrained material. (b) The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.
7.37 In this problem, we are asked for the length of time required for the average grain size of a brass material to increase a specified amount using Figure 7.23. (a) At 500°C, the time necessary for the average grain diameter to increase from 0.01 to 0. mm is approximately 3500 min. (b) At 600°C the time required for this same grain size increase is approximately 150 min.
