# CHAPTER 7 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 7-1 CHAPTER 7 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Basic Concepts of Dislocations Characteristics of Dislocations 7.1 The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic millimeters). Thus, the total length in 1000 mm 3 of material having a density of 10 5 mm - is just (10 5 mm - )(1000 mm 3 ) =10 8 mm = 10 5 m = 6 mi Similarly, for a dislocation density of 10 9 mm -, the total length is (10 9 mm - )(1000 mm 3 ) =10 1 mm = 10 9 m = 6. x 10 5 mi

2 7- dislocations as: 7. When the two edge dislocations become aligned, a planar region of vacancies will exist between the

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.

4 For the various dislocation types, the relationships between the direction of the applied shear stress and the direction of dislocation line motion are as follows: edge dislocation--parallel screw dislocation--perpendicular mixed dislocation--neither parallel nor perpendicular

5 7-5 Slip Systems 7.5 (a) A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs. (b) All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure.

6 (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.11 as PD 110 (FCC) = 1 4 R = R Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.53, which are as follows: PD 100 (FCC) = 1 4 R = 0.5 R PD 111 (FCC) = 1 R 3 = 0.9 R (b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.54, which are as follows: PD 100 (BCC) = 3 16R = 0.19 R PD 110 (BCC) = 3 8 R = 0.7 R Below is a BCC unit cell, within which is shown a (111) plane. (a)

7 7-7 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), the area of this triangle is xy. The triangle edge length, x, is equal to the length of a face diagonal, as indicated in Figure (a). And its length is related to the unit cell edge length, a, as x = a + a = a or x = a For BCC, a = 4 R 3 (Equation 3.3), and, therefore, x = 4R 3 Also, from Figure (b), with respect to the length y we may write

8 7-8 y + x = x which leads to y = x 3. And, substitution for the above expression for x yields y = x 3 = 4 R 3 3 = 4 R Thus, the area of this triangle is equal to AREA = 1 xy= 1 4 R 4 R 3 = 8 R 3 And, finally, the planar density for this (111) plane is 0.5 atom 3 PD 111 (BCC) = 8 R = 16 R = 0.11 R 3

9 Below is shown the atomic packing for a BCC {110}-type plane. The arrows indicate two different <111> type directions.

10 Below is shown the atomic packing for an HCP {0001}-type plane. The arrows indicate three different <11 0 > -type directions.

11 This problem asks that we compute the magnitudes of the Burgers vectors for copper and iron. For Cu, which has an FCC crystal structure, R = nm (Table 3.1) and a = R = nm (Equation 3.1); also, from Equation 7.1a, the Burgers vector for FCC metals is b = a 110 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 u + v + w = nm (1 ) + (1 ) + (0 ) = nm For Fe which has a BCC crystal structure, R = nm (Table 3.1) and a = 4 R 3 = nm (Equation 3.3); also, from Equation 7.1b, the Burgers vector for BCC metals is b = a 111 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 Fe is b = nm (1) + (1) + (1) = 0.48 nm

12 (a) This part of the problem asks that we specify the Burgers vector for the simple cubic crystal structure (and suggests that we consult the answer to Concept Check 7.1). This Concept Check asks that we select the slip system for simple cubic from four possibilities. The correct answer is { 100} 010. Thus, the Burgers vector will lie in a 010 -type direction. Also, the unit slip distance is a (i.e., the unit cell edge length, Figures 4.3 and 7.1). Therefore, the Burgers vector for simple cubic is b = a 010 Or, equivalently b = a 100 (b) The magnitude of the Burgers vector, b, for simple cubic is b = a( ) 1/ = a

13 7-13 Slip in Single Crystals 7.11 We are asked to compute the Schmid factor for an FCC crystal oriented with its [10] direction parallel to the loading axis. With this scheme, slip may occur on the (111) plane and in the [011 ] direction as noted in the figure below. The angle between the [10] and [011 ] directions, λ, may be determined using Equation 7.6 λ=cos 1 u 1 u + v 1 v + w 1 w u 1 + v1 + w1 ( ) u + v + w ( ) where (for [10]) u 1 = 1, v 1 =, w 1 = 0, and (for [011 ]) u = 0, v = 1, w = -1. Therefore, λ is equal to λ=cos 1 (1)(0) + ()(1) + (0)( 1) [(1) + () + (0) ][ (0) + (1) + ( 1) ] = cos 1 10 = 50.8

14 7-14 Now, the angle φ is equal to the angle between the normal to the (111) plane (which is the [111] direction), and the [10] direction. Again from Equation 7.6, and for u 1 = 1, v 1 = 1, w 1 = 1, u = 1, v =, and w = 0, we have φ=cos 1 (1)(1) + (1)() + (1)(0) (1) + (1) + (1) [ ][ (1) + () + (0) ] = cos = 39. Therefore, the Schmid factor is equal to cos λ cos φ = cos(50.8 ) cos(39. ) = 10 3 =

15 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. MPa (900 psi) and 1 MPa (1750 psi), respectively. From Equation 7. τ R = σ cos φ cos λ = (1 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. MPa), the single crystal will not yield. However, from Equation 7.4, the stress at which yielding occurs is σ y = τ crss cos φ cos λ = 6. MPa = 15.1 MPa (cos 60 )(cos 35 ) (00 psi)

16 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. Cosines for the possible λ values are given below. cos(30 ) = 0.87 cos(48 ) = 0.67 cos(78 ) = 0.1 Thus, the slip direction is at an angle of 30 with the tensile axis. (b) From Equation 7.4, the critical resolved shear stress is just τ crss = σ y (cos φ cos λ) max = (.5 MPa) [ cos(65 ) cos(30 ) ] = 0.90 MPa (130 psi)

17 This problem asks that we compute the critical resolved shear stress for nickel. In order to do this, we must employ Equation 7.4, but first it is necessary to solve for the angles λ and φ which are shown in the sketch below. The angle λ is the angle between the tensile axis i.e., along the [001] direction and the slip direction i.e., [1 01]. The angle λ may be determined using Equation 7.6 as λ=cos 1 u 1 u + v 1 v + w 1 w u 1 + v1 + w1 ( ) u + v + w ( ) where (for [001]) u 1 = 0, v 1 = 0, w 1 = 1, and (for [1 01]) u = 1, v = 0, w = 1. Therefore, λ is equal to λ=cos 1 (0)( 1) + (0)(0) + (1)(1) (0) + (0) + (1) [ ][ ( 1) + (0) + (1) ] = cos 1 1 = 45 Furthermore, φ is the angle between the tensile axis the [001] direction and the normal to the slip plane i.e., the (111) plane; for this case this normal is along a [111] direction. Therefore, again using Equation 7.6

18 7-18 φ=cos 1 (0)(1) + (0)(1) + (1)(1) (0) + (0) + (1) [ ][ (1) + (1) + (1) ] = cos = 54.7 And, finally, using Equation 7.4, the critical resolved shear stress is equal to τ crss = σ y (cos φ cos λ) = (13.9 MPa) [ cos(54.7 ) cos(45 ) ]= (13.9 MPa) = 5.68 MPa (85 psi)

19 This problem asks that, for a metal that has the FCC crystal structure, we compute the applied stress(s) that are required to cause slip to occur on a (111) plane in each of the [110 ], [101 ], and [011] directions. In order to solve this problem it is necessary to employ Equation 7.4, but first we need to solve for the for λ and φ angles for the three slip systems. For each of these three slip systems, the φ will be the same i.e., the angle between the direction of the applied stress, [100] and the normal to the (111) plane, that is, the [111] direction. The angle φ may be determined using Equation 7.6 as φ=cos 1 u 1 u + v 1 v + w 1 w u 1 + v1 + w1 ( ) u + v + w ( ) where (for [100]) u 1 = 1, v 1 = 0, w 1 = 0, and (for [111]) u = 1, v = 1, w = 1. Therefore, φ is equal to φ=cos 1 (1)(1) + (0)(1) + (0)(1) (1) + (0) + (0) [ ][ (1) + (1) + (1) ] = cos = 54.7 Let us now determine λ for the [11 0 ] slip direction. Again, using Equation 7.6 where u 1 = 1, v 1 = 0, w 1 = 0 (for [100]), and u = 1, v = 1, w = 0 (for [11 0]. Therefore, λ is determined as λ = cos 1 (1)(1) + (0)( 1) + (0)(0) [100] [110] (1) + (0) + (0) [ ][ (1) + ( 1) + (0) ] = cos 1 1 = 45 Now, we solve for the yield strength for this (111) [110] slip system using Equation 7.4 as σ y = τ crss (cosφ cos λ)

20 7-0 = 0.5 MPa cos (54.7 ) cos (45 ) = 0.5 MPa = 1. MPa (0.578) (0.707) Now, we must determine the value of λ for the (111) [101] slip system that is, the angle between the [100] and [101 ] directions. Again using Equation 7.6 λ = cos 1 [100] [101] (1)(1) + (0)(0) + (0)( 1) [(1) + (0) + (0) ][ (1) + (0) + ( 1) ] = cos 1 1 = 45 Thus, since the values of φ and λ for this (111) [101 ] slip system are the same as for (111) [11 0], so also will σ y be the same viz 1. MPa. And, finally, for the (111) [011] slip system, λ is computed using Equation 7.6 as follows: λ [100] [011] = cos 1 (1)(0) + (0)( 1) + (0)(1) (1) + (0) + (0) [ ][ (0) + ( 1) + (1) ] = cos 1 (0) = 90 Thus, from Equation 7.4, the yield strength for this slip system is σ y = τ crss (cosφ cos λ) = 0.5 MPa 0.5 MPa = cos (54.7 ) cos (90 ) (0.578) (0) = which means that slip will not occur on this (111) [011] slip system.

21 (a) This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the [111] direction on each of the (110), (011), and (10 1 ) planes. In order to solve this problem it is necessary to employ Equation 7., 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, [111]. This angle λ may be determined using Equation 7.6 λ=cos 1 u 1 u + v 1 v + w 1 w u 1 + v1 + w1 ( ) u + v + w ( ) where (for [100]) u 1 = 1, v 1 = 0, w 1 = 0, and (for [11 1]) u = 1, v = 1, w = 1. Therefore, λ is determined as λ=cos 1 (1)(1) + (0)( 1) + (0)(1) (1) + (0) + (0) [ ][ (1) + ( 1) + (1) ] = cos = 54.7 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 7.6 where u 1 = 1, v 1 = 0, w 1 = 0 (for [100]), and u = 1, v = 1, w = 0 (for [110]), φ is equal to φ [100] [110] = cos 1 (1)(1) + (0)(1) + (0)(0) (1) + (0) + (0) [ ][ (1) + (1) + (0) ] = cos 1 1 = 45 Now, using Equation 7. τ R = σ cosφ cos λ we solve for the resolved shear stress for this slip system as

22 7- τ R (110) [11 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) [111] 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 7.6 λ [100] [ 011] = cos 1 (1)(0) + (0)(1) + (0)(1) (1) + (0) + (0) [ ][ (0) + (1) + (1) ] = cos 1 (0) = 90 Thus, the resolved shear stress for this (011) [111] slip system is τ R (011) [11 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 ) [111] slip system that is, the angle between the direction of the applied stress, [100], and the normal to the (101 ) plane i.e., the [101 ] direction. Again using Equation 7.6 λ [100] [101] = cos 1 (1)(1) + (0)(0) + (0)( 1) [(1) + (0) + (0) ][ (1) + (0) + ( 1) ] = cos 1 1 = 45 Here, as with the (110) [111] slip system above, the value of φ is 45, which again leads to τ R (101 ) [11 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 τ R value. Both (110) [11 1] and (101 ) [11 1] slip systems are most favored since they have the same τ R (1.63 MPa), which is greater than the τ R value for (011) [11 1] (viz., 0 MPa).

23 This problem asks for us to determine the tensile stress at which a BCC metal yields when the stress is applied along a [11] direction such that slip occurs on a (101) plane and in a [1 11] direction; the critical resolved shear stress for this metal is.4 MPa. To solve this problem we use Equation 7.4; however it is first necessary to determine the values of φ and λ. These determinations are possible using Equation 7.6. Now, λ is the angle between [11] and [1 11] directions. Therefore, relative to Equation 7.6 let us take u 1 = 1, v 1 =, and w 1 = 1, as well as u = 1, v = 1, and w = 1. This leads to λ=cos 1 u 1 u + v 1 v + w 1 w u 1 + v1 + w1 ( ) u + v + w ( ) = cos 1 (1)( 1) + ()(1) + (1)(1) (1) + () + (1) [ ][ ( 1) + (1) + (1) ] = cos 1 18 = 61.9 Now for the determination of φ, the normal to the (101) slip plane is the [101] direction. Again using Equation 7.6, where we now take u 1 = 1, v 1 =, w 1 = 1 (for [11]), and u = 1, v = 0, w = 1 (for [101]). Thus, φ=cos 1 (1)(1) + ()(0) + (1)(1) (1) + () + (1) [ ][ (1) + (0) + (1) ] = cos 1 1 = 54.7 It is now possible to compute the yield stress (using Equation 7.4) as σ y = τ crss cosφ cos λ =.4 MPa = 8.8 MPa 1 18

24 In order to determine the maximum possible yield strength for a single crystal of Cu pulled in tension, we simply employ Equation 7.5 as σ y =τ crss = ()(0.48 MPa) = 0.96 MPa (140 psi)

25 7-5 Deformation by Twinning 7.19 Four major differences between deformation by twinning and deformation by slip are as follows: (1) with slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation; () for slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples; (3) slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems; and (4) normally slip results in relatively large deformations, whereas only small deformations result for twinning.

26 7-6 Strengthening by Grain Size Reduction 7.0 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 smallangle, and therefore not as much change in slip direction.

27 Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.

28 These three strengthening mechanisms are described in Sections 7.8, 7.9, and 7.10.

29 (a) Perhaps the easiest way to solve for σ 0 and k y in Equation 7.7 is to pick two values each of σ y and d -1/ from Figure 7.15, and then solve two simultaneous equations, which may be set up. For example d -1/ (mm) -1/ σ y (MPa) The two equations are thus 75 = σ 0 + 4k y 175 = σ 0 +1k y Solution of these equations yield the values of k y = 1.5 MPa (mm) 1/ [ 1810 psi(mm) 1/ ] σ 0 = 5 MPa (3630 psi) (b) When d =.0 x 10-3 mm, d -1/ =.4 mm -1/, and, using Equation 7.7, σ y = σ 0 + k y d -1/ = (5 MPa) MPa (mm) 1/ (.4 mm -1/ ) = 305 MPa (44,00 psi)

30 We are asked to determine the grain diameter for an iron which will give a yield strength of 310 MPa (45,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of Equation 7.7, solve for σ 0 and k y, and finally determine the value of d when σ y = 310 MPa. The data pertaining to this problem may be tabulated as follows: σ y d (mm) d -1/ (mm) -1/ 30 MPa 1 x MPa 6 x The two equations thus become 30 MPa = σ 0 + (10.0) k y 75 MPa = σ 0 + (1.91) k y Which yield the values, σ 0 = 75.4 MPa and k y = MPa(mm) 1/. At a yield strength of 310 MPa 310 MPa = 75.4 MPa + [ MPa (mm) 1/ ]d -1/ or d -1/ = (mm) -1/, which gives d = 4.34 x 10-3 mm.

31 This problem asks that we determine the grain size of the brass for which is the subject of Figure From Figure 7.19(a), the yield strength of brass at 0%CW is approximately 175 MPa (6,000 psi). This yield strength from Figure 7.15 corresponds to a d -1/ value of approximately 1.0 (mm) -1/. Thus, d = 6.9 x 10-3 mm.

32 7-3 Solid-Solution Strengthening 7.6 Below is shown an edge dislocation and where an interstitial impurity atom would be located. Compressive lattice strains are introduced by the impurity atom. There will be a net reduction in lattice strain energy when these lattice strains partially cancel tensile strains associated with the edge dislocation; such tensile strains exist just below the bottom of the extra half-plane of atoms (Figure 7.4).

33 7-33 Strain Hardening 7.7 (a) We are asked to show, for a tensile test, that ε %CW = x 100 ε+1 From Equation 7.8 %CW = A 0 A d A 0 x 100 = 1 A d A 0 x 100 Which is also equal to 1 l 0 l d x 100 since A d /A 0 = l 0 /l d, the conservation of volume stipulation given in the problem statement. Now, from the definition of engineering strain (Equation 6.) ε = l d l 0 l 0 = l d l 0 1 Or, l 0 l d = 1 ε+1 Substitution for l 0 / l d into the %CW expression above gives %CW = 1 l 0 l d x 100 = 1 1 ε x 100 = x 100 ε+1 ε+1 expression (b) From Figure 6.1, a stress of 415 MPa (60,000 psi) corresponds to a strain of Using the above ε 0.16 %CW = x 100 = x 100 = 13.8%CW ε

34 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 %CW = A 0 A d x 100 = π r 0 πrd A 0 π r 0 x 100 = π (15 mm) π(1 mm) π (15 mm) x 100 = 36%CW For the second specimen, the deformed radius is computed using the above equation and solving for r d as r d = r 0 1 %CW 100 = (11 mm) 1 36%CW 100 = 8.80 mm

35 We are given the original and deformed cross-sectional dimensions for two specimens of the same metal, and are then asked to determine which is the hardest after deformation. 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 %CW = A 0 A d A 0 x 100 = π r 0 πr d π r 0 x mm 15.9 mm π π = 18.0 mm π x 100 =.0%CW For the rectangular one (0 mm)(50 mm) (13.7 mm)(55.1 mm) %CW = x 100 = 4.5%CW (0 mm)(50 mm) Therefore, the deformed rectangular specimen will be harder.

36 This problem calls for us to calculate the precold-worked radius of a cylindrical specimen of copper that has a cold-worked ductility of 15%EL. From Figure 7.19(c), copper that has a ductility of 15%EL will have experienced a deformation of about 0%CW. For a cylindrical specimen, Equation 7.8 becomes %CW = π r 0 πr d π r 0 x 100 Since r d = 6.4 mm (0.5 in.), solving for r 0 yields r 0 = r d 1 %CW 100 = 6.4 mm = 7. mm (0.80 in.)

37 (a) We want to compute the ductility of a brass that has a yield strength of 345 MPa (50,000 psi). In order to solve this problem, it is necessary to consult Figures 7.19(a) and (c). From Figure 7.19(a), a yield strength of 345 MPa for brass corresponds to 0%CW. A brass that has been cold-worked 0% will have a ductility of about 4%EL [Figure 7.19(c)]. (b) This portion of the problem asks for the Brinell hardness of a 1040 steel having a yield strength of 60 MPa (90,000 psi). From Figure 7.19(a), a yield strength of 60 MPa for a 1040 steel corresponds to about 5%CW. A 1040 steel that has been cold worked 5% will have a tensile strength of about 750 MPa [Figure 7.19(b)]. Finally, using Equation 6.0a HB = TS (MPa) 3.45 = 750 MPa 3.45 = 17

38 We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 10 6 mm -. It is first necessary to compute the value of the constant A (in the equation provided in the problem statement) from the one set of data as A = τ crss τ 0 ρ D = 0.69 MPa MPa 10 4 mm = 6.1 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 - as τ crss = τ 0 + A ρ D = (0.069 MPa) + (6.1 x 10-3 MPa - mm) 10 6 mm = 6.8 MPa (910 psi)

39 7-39 Recovery Recrystallization Grain Growth 7.33 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.

40 We are asked to estimate the fraction of recrystallization from the photomicrograph in Figure 7.1c. Below is shown a square grid onto which is superimposed the recrystallized regions from the micrograph. Approximately 400 squares lie within the recrystallized areas, and since there are 67 total squares, the specimen is about 60% recrystallized.

41 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.

42 (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.

43 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 7.5. (a) At 600 C, the time necessary for the average grain diameter to grow to 0.03 is about 6 min; and the total time to grow to 0.3 mm is approximately 3000 min. Therefore, the time to grow from 0.03 to 0.3 mm is 3000 min - 6 min, or approximately 3000 min. (b) At 700 C the time required for this same grain size increase is approximately 80 min.

44 (a) Using the data given and Equation 7.9 (taking n = ), we may set up two simultaneous equations with d 0 and K as unknowns; thus (5.6 x 10 - mm) d 0 = (40 min)k (8.0 x 10 - mm) d 0 = (100 min)k Solution of these expressions yields a value for d 0, the original grain diameter, of and a value for K of 5.44 x 10-5 mm /min d 0 = mm, (b) At 00 min, the diameter d is computed using a rearranged form of Equation 7.9 as d = d 0 + Kt = (0.031 mm) + (5.44 x 10 5 mm /min)(00 min) = mm

45 Yes, it is possible to reduce the average grain diameter of an undeformed alloy specimen from mm to 0.00 mm. In order to do this, plastically deform the material at room temperature (i.e., cold work it), and then anneal at an elevated temperature in order to allow recrystallization and some grain growth to occur until the average grain diameter is 0.00 mm.

46 (a) The temperature dependence of grain growth is incorporated into the constant K in Equation 7.9. (b) The explicit expression for this temperature dependence is of the form K = K 0 exp Q RT in which K 0 is a temperature-independent constant, the parameter Q is an activation energy, and R and T are the gas constant and absolute temperature, respectively.

47 This problem calls for us to calculate the yield strength of a brass specimen after it has been heated to an elevated temperature at which grain growth was allowed to occur; the yield strength (150 MPa) was given at a grain size of 0.01 mm. It is first necessary to calculate the constant k y in Equation 7.7 as k y = σ y σ 0 d -1/ = 150 MPa 5 MPa (0.01 mm) 1/ = 1.5 MPa mm 1/ Next, we must determine the average grain size after the heat treatment. From Figure 7.5 at 500 C after 1000 s (16.7 min) the average grain size of a brass material is about mm. Therefore, calculating σ y at this new grain size using Equation 7.7 we get σ y = σ 0 + k y d -1/ = 5 MPa + (1.5 MPa - mm 1/ )(0.016 mm) -1/ = 14 MPa (18,000 psi)

48 7-48 DESIGN PROBLEMS Strain Hardening Recrystallization 7.D1 This problem calls for us to determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 40 and a ductility of at least 15%EL. According to Figure 6.19, a Brinell hardness of 40 corresponds to a tensile strength of 800 MPa (116,000 psi). Furthermore, from Figure 7.19(b), in order to achieve a tensile strength of 800 MPa, deformation of at least 13%CW is necessary. Finally, if we cold work the steel to 13%CW, then the ductility is 15%EL from Figure 7.19(c). Therefore, it is possible to meet both of these criteria by plastically deforming the steel.

49 D We are asked to determine whether or not it is possible to cold work brass so as to give a minimum Brinell hardness of 150 and at the same time have a ductility of at least 0%EL. According to Figure 6.19, a Brinell hardness of 150 corresponds to a tensile strength of 500 MPa (7,000 psi.) Furthermore, from Figure 7.19(b), in order to achieve a tensile strength of 500 MPa, deformation of at least 36%CW is necessary. Finally, if we are to achieve a ductility of at least 0%EL, then a maximum deformation of 3%CW is possible from Figure 7.19(c). Therefore, it is not possible to meet both of these criteria by plastically deforming brass.

50 D3 (a) For this portion of the problem we are to determine the ductility of cold-worked steel that has a Brinell hardness of 40. From Figure 6.19, a Brinell hardness of 40 corresponds to a tensile strength of 80 MPa (10,000 psi), which, from Figure 7.19(b), requires a deformation of 17%CW. Furthermore, 17%CW yields a ductility of about 13%EL for steel, Figure 7.19(c). (b) We are now asked to determine the radius after deformation if the uncold-worked radius is 10 mm (0.40 in.). From Equation 7.8 and for a cylindrical specimen %CW = π r 0 πr d π r 0 x 100 Now, solving for r d from this expression, we get r d = r 0 1 %CW 100 = (10 mm) = 9.11 mm (0.364 in.)

51 D4 This problem asks us to determine which of copper, brass, and a 1040 steel may be cold-worked so as to achieve a minimum yield strength of 310 MPa (45,000 psi) while maintaining a minimum ductility of 7%EL. For each of these alloys, the minimum cold work necessary to achieve the yield strength may be determined from Figure 7.19(a), while the maximum possible cold work for the ductility is found in Figure 7.19(c). These data are tabulated below. Yield Strength Ductility (> 310 MPa) (> 7%EL) Steel Any %CW Not possible Brass > 15%CW < 18%CW Copper > 38%CW < 10%CW Thus, only brass is a possible candidate since for this alloy only there is an overlap of %CW's to give the required minimum yield strength and ductility values.

52 7-5 7.D5 This problem calls for us to explain the procedure by which a cylindrical rod of 1040 steel may be deformed so as to produce a given final diameter (8.9 mm), as well as a specific minimum tensile strength (85 MPa) and minimum ductility (1%EL). First let us calculate the percent cold work and attendant tensile strength and ductility if the drawing is carried out without interruption. From Equation 7.8 %CW = π d 0 π d d π d 0 x 100 = 11.4 mm 8.9 mm π π 11.4 mm π x 100 = 40%CW At 40%CW, the steel will have a tensile strength on the order of 900 MPa (130,000 psi) [Figure 7.19(b)], which is adequate; however, the ductility will be less than 9%EL [Figure 7.19(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 7.19(b) indicates that 17%CW is necessary to yield a tensile strength of 85 MPa (1,000 psi). Similarly, a maximum of 19%CW is possible for 1%EL [Figure 7.19(c)]. The average of these extremes is 18%CW. If the final diameter after the first drawing is d ' 0, then 18%CW = π d 0 ' π π d 0 ' 8.9 mm x 100 And, solving for d 0 ', yields d 0 ' = 8.9 mm 1 18%CW 100 = 9.83 mm (0.387 in.)

53 D6 Let us first calculate the percent cold work and attendant yield strength and ductility if the drawing is carried out without interruption. From Equation 7.8 %CW = π d 0 π d d π d 0 x 100 = 10. mm 7.6 mm π π 10. mm π x 100 = 44.5%CW At 44.5%CW, the brass will have a yield strength on the order of 40 MPa (61,000 psi), Figure 7.19(a), which is adequate; however, the ductility will be about 5%EL, Figure 7.19(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, yield strength, and ductility. Reference to Figure 7.19(a) indicates that 7.5%CW is necessary to give a yield strength of 380 MPa. Similarly, a maximum of 7.5%CW is possible for 15%EL [Figure 7.19(c)]. Thus, to achieve both the specified yield strength and ductility, the brass must be deformed to 7.5 %CW. If the final diameter after the first drawing is d ' 0, then, using Equation %CW = π d 0 ' π π d 0 ' 7.6 mm x 100 And, solving for d 0 ' yields d 0 ' = 7.6 mm 1 7.5%CW 100 = 8.93 mm (0.351 in.)

54 D7 This problem calls for us to cold work some brass stock that has been previously cold worked in order to achieve minimum tensile strength and ductility values of 450 MPa (65,000 psi) and 13%EL, respectively, while the final diameter must be 1.7 mm (0.50 in.). Furthermore, the material may not be deformed beyond 65%CW. Let us start by deciding what percent coldwork is necessary for the minimum tensile strength and ductility values, assuming that a recrystallization heat treatment is possible. From Figure 7.19(b), at least 7%CW is required for a tensile strength of 450 MPa. Furthermore, according to Figure 7.19(c), 13%EL corresponds a maximum of 30%CW. Let us take the average of these two values (i.e., 8.5%CW), and determine what previous specimen diameter is required to yield a final diameter of 1.7 mm. For cylindrical specimens, Equation 7.8 takes the form %CW = π d 0 π d d π d 0 x 100 Solving for the original diameter d 0 yields d 0 = d d 1 %CW 100 = 1.7 mm = 15.0 mm (0.591 in.) Now, let us determine its undeformed diameter realizing that a diameter of 19.0 mm corresponds to 35%CW. Again solving for d 0 using the above equation and assuming d d = 19.0 mm yields d 0 = d d 1 %CW 100 = 19.0 mm = 3.6 mm (0.930 in.) At this point let us see if it is possible to deform the material from 3.6 mm to 15.0 mm without exceeding the 65%CW limit. Again employing Equation 7.8 %CW = 3.6 mm 15.0 mm π π 3.6 mm π x 100 = 59.6%CW

55 7-55 In summary, the procedure which can be used to produce the desired material would be as follows: cold work the as-received stock to 15.0 mm (0.591 in.), heat treat it to achieve complete recrystallization, and then cold work the material again to 1.7 mm (0.50 in.), which will give the desired tensile strength and ductility.

### (10 4 mm -2 )(1000 mm 3 ) = 10 7 mm = 10 4 m = 6.2 mi

14:440:407 Fall 010 Additional problems and SOLUTION OF HOMEWORK 07 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Basic Concepts of Dislocations Characteristics of Dislocations 7.1 To provide

### Concepts of Stress and Strain

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 10 6 psi) and an original

### (10 4 mm -2 )(1000 mm 3 ) = 10 7 mm = 10 4 m = 6.2 mi

CHAPTER 8 DEFORMATION AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Basic Concepts of Dislocations Characteristics of Dislocations 8.1 To provide some perspective on the dimensions of atomic defects,

### Chapter Outline Dislocations and Strengthening Mechanisms

Chapter Outline Dislocations and Strengthening Mechanisms What is happening in material during plastic deformation? Dislocations and Plastic Deformation Motion of dislocations in response to stress Slip

### Chapter Outline Dislocations and Strengthening Mechanisms

Chapter Outline Dislocations and Strengthening Mechanisms What is happening in material during plastic deformation? Dislocations and Plastic Deformation Motion of dislocations in response to stress Slip

### For Cu, which has an FCC crystal structure, R = nm (Table 3.1) and a = 2R 2 =

7.9 Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crstal structures, are of the form a b = uvw where a is the unit cell edge length. Also, since the magnitudes of these Burgers

### CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS

4-1 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation

### Strengthening. Mechanisms of strengthening in single-phase metals: grain-size reduction solid-solution alloying strain hardening

Strengthening The ability of a metal to deform depends on the ability of dislocations to move Restricting dislocation motion makes the material stronger Mechanisms of strengthening in single-phase metals:

### Material Deformations. Academic Resource Center

Material Deformations Academic Resource Center Agenda Origin of deformations Deformations & dislocations Dislocation motion Slip systems Stresses involved with deformation Deformation by twinning Origin

### CHAPTER 7: DISLOCATIONS AND STRENGTHENING

CHAPTER 7: DISLOCATIONS AND STRENGTHENING ISSUES TO ADDRESS... Why are dislocations observed primarily in metals and alloys? Mech 221 - Notes 7 1 DISLOCATION MOTION Produces plastic deformation, in crystalline

### LECTURE SUMMARY September 30th 2009

LECTURE SUMMARY September 30 th 2009 Key Lecture Topics Crystal Structures in Relation to Slip Systems Resolved Shear Stress Using a Stereographic Projection to Determine the Active Slip System Slip Planes

### Mechanical Properties of Metals Mechanical Properties refers to the behavior of material when external forces are applied

Mechanical Properties of Metals Mechanical Properties refers to the behavior of material when external forces are applied Stress and strain fracture or engineering point of view: allows to predict the

### Material Strengthening Mechanisms. Academic Resource Center

Material Strengthening Mechanisms Academic Resource Center Agenda Definition of strengthening Strengthening mechanisms Grain size reduction Solid solution alloying Cold Working (strain hardening) Three

### 14:635:407:02 Homework III Solutions

14:635:407:0 Homework III Solutions 4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 37 C (600 K). Assume an energy for vacancy formation of 0.55 ev/atom.

### Lösungen Übung Verformung

Lösungen Übung Verformung 1. (a) What is the meaning of T G? (b) To which materials does it apply? (c) What effect does it have on the toughness and on the stress- strain diagram? 2. Name the four main

### Chapter Outline. Mechanical Properties of Metals How do metals respond to external loads?

Mechanical Properties of Metals How do metals respond to external loads? Stress and Strain Tension Compression Shear Torsion Elastic deformation Plastic Deformation Yield Strength Tensile Strength Ductility

### The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C. = 2(sphere volume) = 2 = V C = 4R

3.5 Show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated

### Question 6.5: A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm

14:440:407 Ch6 Question 6.3: A specimen of aluminum having a rectangular cross section 10 mm 12.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (8000 lb f ) force, producing only elastic deformation.

### Tensile Testing. Objectives

Laboratory 1 Tensile Testing Objectives Students are required to understand the principle of a uniaxial tensile testing and gain their practices on operating the tensile testing machine. Students are able

### Experiment: Crystal Structure Analysis in Engineering Materials

Experiment: Crystal Structure Analysis in Engineering Materials Objective The purpose of this experiment is to introduce students to the use of X-ray diffraction techniques for investigating various types

### MAE 20 Winter 2011 Assignment 2 solutions

MAE 0 Winter 0 Assignment solutions. List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure (Figure.6). In Figure.6, the barium ions are

### CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram),

### Lecture 18 Strain Hardening And Recrystallization

-138- Lecture 18 Strain Hardening And Recrystallization Strain Hardening We have previously seen that the flow stress (the stress necessary to produce a certain plastic strain rate) increases with increasing

### Energy of a Dislocation

Energy of a Dislocation The Line Tension T Broken and stretched bonds around the dislocation There is EXTRA energy associated with the Defect T = G 2 r 2 b J m G = Shear Modulus Total Extra Energy in the

### x100 A o Percent cold work = %CW = A o A d Yield Stress Work Hardening Why? Cell Structures Pattern Formation

Work Hardening Dislocations interact with each other and assume configurations that restrict the movement of other dislocations. As the dislocation density increases there is an increase in the flow stress

### Solution for Homework #1

Solution for Homework #1 Chapter 2: Multiple Choice Questions (2.5, 2.6, 2.8, 2.11) 2.5 Which of the following bond types are classified as primary bonds (more than one)? (a) covalent bonding, (b) hydrogen

### Deformation of Single Crystals

Deformation of Single Crystals When a single crystal is deformed under a tensile stress, it is observed that plastic deformation occurs by slip on well defined parallel crystal planes. Sections of the

### σ = F / A o Chapter Outline Introduction Mechanical Properties of Metals How do metals respond to external loads?

Mechanical Properties of Metals How do metals respond to external loads? and Tension Compression Shear Torsion Elastic deformation Chapter Outline Introduction To understand and describe how materials

### Relevant Reading for this Lecture... Pages 83-87.

LECTURE #06 Chapter 3: X-ray Diffraction and Crystal Structure Determination Learning Objectives To describe crystals in terms of the stacking of planes. How to use a dot product to solve for the angles

### MAE 20 Winter 2011 Assignment 3 solutions

MAE 20 Winter 2011 Assignment 3 solutions 4.3 Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500 C (773 K) is 7.57 10 23 m -3. The

### Materials Science and Engineering Department MSE , Sample Test #1, Spring 2010

Materials Science and Engineering Department MSE 200-001, Sample Test #1, Spring 2010 ID number First letter of your last name: Name: No notes, books, or information stored in calculator memories may be

### CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS

CHAPTER THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts.6 Show that the atomic packing factor for HCP is 0.74. The APF is just the total sphere volume-unit cell volume ratio.

### ME 612 Metal Forming and Theory of Plasticity. 1. Introduction

Metal Forming and Theory of Plasticity Yrd.Doç. e mail: azsenalp@gyte.edu.tr Makine Mühendisliği Bölümü Gebze Yüksek Teknoloji Enstitüsü In general, it is possible to evaluate metal forming operations

### Defects Introduction. Bonding + Structure + Defects. Properties

Defects Introduction Bonding + Structure + Defects Properties The processing determines the defects Composition Bonding type Structure of Crystalline Processing factors Defects Microstructure Types of

### Solution: ΔG = a 3 ΔG v. + 6a 2 γ. Differentiation of this expression with respect to a is as. d ΔG da = d(a3 ΔG v. + d(6a2 γ) = 3a 2 ΔG v.

14:440:407 ch10 Question 10.2: (a) Rewrite the expression for the total free energy change for nucleation (Equation 10.1) for the case of a cubic nucleus of edge length a (instead of a sphere of radius

### Dislocation theory. Subjects of interest

Chapter 5 Dislocation theory Subjects of interest Introduction/Objectives Observation of dislocation Burgers vector and the dislocation loop Dislocation in the FCC, HCP and BCC lattice Stress fields and

### v kt = N A ρ Au exp (

4-2 4.2 Determination of the number of vacancies per cubic meter in gold at 900 C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v N exp Q v N A ρ Au kt A Au exp Q v kt (6.023

### MAE 20 Winter 2011 Assignment 5

MAE 20 Winter 2011 Assignment 5 6.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 10 6 psi). (a) What is the

### Module #17. Work/Strain Hardening. READING LIST DIETER: Ch. 4, pp. 138-143; Ch. 6 in Dieter

Module #17 Work/Strain Hardening READING LIST DIETER: Ch. 4, pp. 138-143; Ch. 6 in Dieter D. Kuhlmann-Wilsdorf, Trans. AIME, v. 224 (1962) pp. 1047-1061 Work Hardening RECALL: During plastic deformation,

### Torsion Testing. Objectives

Laboratory 4 Torsion Testing Objectives Students are required to understand the principles of torsion testing, practice their testing skills and interpreting the experimental results of the provided materials

### Chapter Outline: Phase Transformations in Metals

Chapter Outline: Phase Transformations in Metals Heat Treatment (time and temperature) Microstructure Mechanical Properties Kinetics of phase transformations Multiphase Transformations Phase transformations

### Chapter Outline. How do atoms arrange themselves to form solids?

Chapter Outline How do atoms arrange themselves to form solids? Fundamental concepts and language Unit cells Crystal structures Simple cubic Face-centered cubic Body-centered cubic Hexagonal close-packed

### Chapter Outline. Diffusion - how do atoms move through solids?

Chapter Outline iffusion - how do atoms move through solids? iffusion mechanisms Vacancy diffusion Interstitial diffusion Impurities The mathematics of diffusion Steady-state diffusion (Fick s first law)

### Stress Strain Relationships

Stress Strain Relationships Tensile Testing One basic ingredient in the study of the mechanics of deformable bodies is the resistive properties of materials. These properties relate the stresses to the

### CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS. C Zn A Cu C Zn A Cu C Cu A Zn. (30)(63.55 g /mol) (30)(63.55 g /mol) (70)(65.

HAPTER 4 IMPERFETIONS IN SOLIDS PROBLEM SOLUTIONS wt% u? 4.7 What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 4.6 as In order to compute composition, in atom percent,

CH 6: Fatigue Failure Resulting from Variable Loading Some machine elements are subjected to static loads and for such elements static failure theories are used to predict failure (yielding or fracture).

Problem Set 7 Materials101 1.) You wish to develop a gold alloy (mostly gold) that can be precipitation strengthened to provide high strength - high conductivity electrical leads to integrated circuits

### Fatigue :Failure under fluctuating / cyclic stress

Fatigue :Failure under fluctuating / cyclic stress Under fluctuating / cyclic stresses, failure can occur at loads considerably lower than tensile or yield strengths of material under a static load: Fatigue

### Chapter Outline. Mechanical Properties of Metals How do metals respond to external loads?

Mechanical Properties of Metals How do metals respond to external loads? Stress and Strain Tension Compression Shear Torsion Elastic deformation Plastic Deformation Yield Strength Tensile Strength Ductility

### Imperfections in atomic arrangements

MME131: Lecture 8 Imperfections in atomic arrangements Part 1: 0D Defects A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Occurrence and importance of crystal defects Classification

### Composition Solidus Liquidus (wt% Au) Temperature ( C) Temperature ( C)

10.4 Given here are the solidus and liquidus temperatures for the copper gold system. Construct the phase diagram for this system and label each region. Composition Solidus Liquidus (wt% Au) Temperature

### Met-2023: Concepts of Materials Science I Sample Questions & Answers,(2009) ( Met, PR, FC, MP, CNC, McE )

1 Met-223: Concepts of Materials Science I Sample Questions & Answers,(29) ( Met, PR, FC, MP, CNC, McE ) Q-1.Define the following. (i) Point Defects (ii) Burgers Vector (iii) Slip and Slip system (iv)

### Principles of Fracture Mechanics

8-1 CHAPTER 8 FAILURE PROBLEM SOLUTIONS Principles of Fracture Mechanics 8.1 This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation

### Welcome to the first lesson of third module which is on thin-walled pressure vessels part one which is on the application of stress and strain.

Strength of Materials Prof S. K. Bhattacharya Department of Civil Engineering Indian Institute of Technology, Kharagpur Lecture -15 Application of Stress by Strain Thin-walled Pressure Vessels - I Welcome

### Fatigue Testing. Objectives

Laboratory 8 Fatigue Testing Objectives Students are required to understand principle of fatigue testing as well as practice how to operate the fatigue testing machine in a reverse loading manner. Students

### Chapter 5: Diffusion. 5.1 Steady-State Diffusion

: Diffusion Diffusion: the movement of particles in a solid from an area of high concentration to an area of low concentration, resulting in the uniform distribution of the substance Diffusion is process

### Constitutive Equations - Plasticity

MCEN 5023/ASEN 5012 Chapter 9 Constitutive Equations - Plasticity Fall, 2006 1 Mechanical Properties of Materials: Modulus of Elasticity Tensile strength Yield Strength Compressive strength Hardness Impact

### The University of Western Ontario Department of Physics and Astronomy P2800 Fall 2008

P2800 Fall 2008 Questions (Total - 20 points): 1. Of the noble gases Ne, Ar, Kr and Xe, which should be the most chemically reactive and why? (0.5 point) Xenon should be most reactive since its outermost

### Materials Issues in Fatigue and Fracture

Materials Issues in Fatigue and Fracture 5.1 Fundamental Concepts 5.2 Ensuring Infinite Life 5.3 Finite Life 5.4 Summary FCP 1 5.1 Fundamental Concepts Structural metals Process of fatigue A simple view

### Chapter 3: The Structure of Crystalline Solids

Sapphire: cryst. Al 2 O 3 Insulin : The Structure of Crystalline Solids Crystal: a solid composed of atoms, ions, or molecules arranged in a pattern that is repeated in three dimensions A material in which

### Basic Concepts of Crystallography

Basic Concepts of Crystallography Language of Crystallography: Real Space Combination of local (point) symmetry elements, which include angular rotation, center-symmetric inversion, and reflection in mirror

### Lecture 14. Chapter 8-1

Lecture 14 Fatigue & Creep in Engineering Materials (Chapter 8) Chapter 8-1 Fatigue Fatigue = failure under applied cyclic stress. specimen compression on top bearing bearing motor counter flex coupling

### Dislocation Plasticity: Overview

Dislocation Plasticity: Overview 1. DISLOCATIONS AND PLASTIC DEFORMATION An arbitrary deformation of a material can always be described as the sum of a change in volume and a change in shape at constant

### Torsion Tests. Subjects of interest

Chapter 10 Torsion Tests Subjects of interest Introduction/Objectives Mechanical properties in torsion Torsional stresses for large plastic strains Type of torsion failures Torsion test vs.tension test

### Iron-Carbon Phase Diagram (a review) see Callister Chapter 9

Iron-Carbon Phase Diagram (a review) see Callister Chapter 9 University of Tennessee, Dept. of Materials Science and Engineering 1 The Iron Iron Carbide (Fe Fe 3 C) Phase Diagram In their simplest form,

### Griffith theory of brittle fracture:

Griffith theory of brittle fracture: Observed fracture strength is always lower than theoretical cohesive strength. Griffith explained that the discrepancy is due to the inherent defects in brittle materials

### Mechanical Properties - Stresses & Strains

Mechanical Properties - Stresses & Strains Types of Deformation : Elasic Plastic Anelastic Elastic deformation is defined as instantaneous recoverable deformation Hooke's law : For tensile loading, σ =

### Yield Criteria for Ductile Materials and Fracture Mechanics of Brittle Materials. τ xy 2σ y. σ x 3. τ yz 2σ z 3. ) 2 + ( σ 3. σ 3

Yield Criteria for Ductile Materials and Fracture Mechanics of Brittle Materials Brittle materials are materials that display Hookean behavior (linear relationship between stress and strain) and which

### In order to solve this problem it is first necessary to use Equation 5.5: x 2 Dt. = 1 erf. = 1.30, and x = 2 mm = 2 10-3 m. Thus,

5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Solution (a) With vacancy diffusion,

### Introduction to Materials Science, Chapter 11, Thermal Processing of Metal Alloys. Chapter 11 Thermal Processing of Metal Alloys

Chapter 11 Thermal Processing of Metal Alloys Designer Alloys: Utilize heat treatments to design optimum microstructures and mechanical properties (strength, ductility, hardness.) Strength in steels correlates

### MATERIALS SCIENCE AND ENGINEERING

Complete Solutions to Selected Problems to accompany MATERIALS SCIENCE AND ENGINEERING AN INTRODUCTION Sixth Edition William D. Callister, Jr. The University of Utah John Wiley & Sons, Inc. Copyright 2003

### Martensite in Steels

Materials Science & Metallurgy http://www.msm.cam.ac.uk/phase-trans/2002/martensite.html H. K. D. H. Bhadeshia Martensite in Steels The name martensite is after the German scientist Martens. It was used

### Crystal Symmetries METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Crystal Symmetries Why should we be interested? Important physical properties depend on crystal structure Conductivity Magnetic properties Stiffness Strength These properties also often depend on crystal

### Plastic Behaviour - Tensile Strength

Plastic Behaviour - Tensile Strength Transition of mechanical behaviour from elastic to plastic depends upon the material type and its condition as tested (hot-rolled, cold-rolled, heat treated, etc.).

### Uniaxial Tension and Compression Testing of Materials. Nikita Khlystov Daniel Lizardo Keisuke Matsushita Jennie Zheng

Uniaxial Tension and Compression Testing of Materials Nikita Khlystov Daniel Lizardo Keisuke Matsushita Jennie Zheng 3.032 Lab Report September 25, 2013 I. Introduction Understanding material mechanics

### THE WAY TO SOMEWHERE. Sub-topics. Diffusion Diffusion processes in industry

THE WAY TO SOMEWHERE Sub-topics 1 Diffusion Diffusion processes in industry RATE PROCESSES IN SOLIDS At any temperature different from absolute zero all atoms, irrespective of their state of aggregation

### Hardness Test. Subjects of interest

Hardness Test Chapter 9 Subjects of interest Introduction/objectives Brinell hardness Meyer hardness Vickers hardness Rockwell hardness Microhardness tests Relationship between hardness and the flow curve

### ME 215 Engineering Materials I

ME 215 Engineering Materials I Chapter 3 Properties in Tension and Compression (Part III) Mechanical Engineering University of Gaziantep Dr. A. Tolga Bozdana www.gantep.edu.tr/~bozdana True Stress and

### METU DEPARTMENT OF METALLURGICAL AND MATERIALS ENGINEERING

METU DEPARTMENT OF METALLURGICAL AND MATERIALS ENGINEERING Met E 206 MATERIALS LABORATORY EXPERIMENT 1 Prof. Dr. Rıza GÜRBÜZ Res. Assist. Gül ÇEVİK (Room: B-306) INTRODUCTION TENSION TEST Mechanical testing

### Solid Mechanics. Stress. What you ll learn: Motivation

Solid Mechanics Stress What you ll learn: What is stress? Why stress is important? What are normal and shear stresses? What is strain? Hooke s law (relationship between stress and strain) Stress strain

### Crystal Defects p. 2. Point Defects: Vacancies. Department of Materials Science and Engineering University of Virginia. Lecturer: Leonid V.

Crystal Defects p. 1 A two-dimensional representation of a perfect single crystal with regular arrangement of atoms. But nothing is perfect, and structures of real materials can be better represented by

### Mechanical properties 3 Bend test and hardness test of materials

MME131: Lecture 15 Mechanical properties 3 Bend test and hardness test of materials A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Bend test of brittle materials Hardness testing

### COMPLEX STRESS TUTORIAL 3 COMPLEX STRESS AND STRAIN

COMPLX STRSS TUTORIAL COMPLX STRSS AND STRAIN This tutorial is not part of the decel unit mechanical Principles but covers elements of the following sllabi. o Parts of the ngineering Council eam subject

### Analysis of Stresses and Strains

Chapter 7 Analysis of Stresses and Strains 7.1 Introduction axial load = P / A torsional load in circular shaft = T / I p bending moment and shear force in beam = M y / I = V Q / I b in this chapter, we

### Objectives. Experimentally determine the yield strength, tensile strength, and modules of elasticity and ductility of given materials.

Lab 3 Tension Test Objectives Concepts Background Experimental Procedure Report Requirements Discussion Objectives Experimentally determine the yield strength, tensile strength, and modules of elasticity

### PLASTIC DEFORMATION AND STRESS-STRAIN CURVES

PLASTIC DEFORMATION AND STRESS-STRAIN CURVES Introduction Background Unit: Plastic Deformation and Stress-Strain Curves In the last unit we studied the elastic response of materials to externally applied

### Force on a square loop of current in a uniform B-field.

Force on a square loop of current in a uniform B-field. F top = 0 θ = 0; sinθ = 0; so F B = 0 F bottom = 0 F left = I a B (out of page) F right = I a B (into page) Assume loop is on a frictionless axis

### CHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS

19-1 CHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS Heat Capacity 19.1 The energy, E, required to raise the temperature of a given mass of material, m, is the product of the specific heat, the mass of

### EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS

ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain,

### Ch. 4: Imperfections in Solids Part 1. Dr. Feras Fraige

Ch. 4: Imperfections in Solids Part 1 Dr. Feras Fraige Outline Defects in Solids 0D, Point defects vacancies Interstitials impurities, weight and atomic composition 1D, Dislocations edge screw 2D, Grain

### Mechanical Properties

Mechanical Properties Hardness Hardness can be defined as resistance to deformation or indentation or resistance to scratch. Hardness Indentation Scratch Rebound Indentation hardness is of particular interest

### Chapter Outline. Defects Introduction (I)

Crystals are like people, it is the defects in them which tend to make them interesting! - Colin Humphreys. Defects in Solids Chapter Outline 0D, Point defects vacancies interstitials impurities, weight

### MAE 20 Winter 2011 Assignment 6

MAE 0 Winter 011 Assignment 6 8.3 If the specific surface energy for soda-lime glass is 0.30 J/m, using data contained in Table 1.5, compute the critical stress required for the propagation of a surface

### Figure 1.1 Vector A and Vector F

CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have

### , to obtain a way to calculate stress from the energy function U(r).

BIOEN 36 013 LECTURE : MOLECULAR BASIS OF ELASTICITY Estimating Young s Modulus from Bond Energies and Structures First we consider solids, which include mostly nonbiological materials, such as metals,

### Fundamentals of grain boundaries and grain boundary migration

1. Fundamentals of grain boundaries and grain boundary migration 1.1. Introduction The properties of crystalline metallic materials are determined by their deviation from a perfect crystal lattice, which

### ORIENTATION CHARACTERISTICS OF THE MICROSTRUCTURE OF MATERIALS

ORIENTATION CHARACTERISTICS OF THE MICROSTRUCTURE OF MATERIALS K. Sztwiertnia Polish Academy of Sciences, Institute of Metallurgy and Materials Science, 25 Reymonta St., 30-059 Krakow, Poland MMN 2009

### MATERIALS SCIENCE AND ENGINEERING Vol. II Structural and Functional Materials - H. V. Atkinson STRUCTURAL AND FUNCTIONAL MATERIALS

STRUCTURAL AND FUNCTIONAL MATERIALS H. V. Atkinson University of Sheffield, UK Keywords: Structural materials, functional materials, optical, electrical, dielectric, magnetic, thermal, elastic modulus,