# For each of edge screw and mixed dislocations cite the relationship

### For each of edge, screw, and mixed dislocations, cite the | StudySoup

4 For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line. Summary: For an edge dislocation, the Burgers vector is parallel to dislocation motion. For a screw dislocation, the Burgers vector is parallel to. In materials science, a dislocation or Taylor's dislocation is a crystallographic defect or The two primary types of dislocations are edge dislocations and screw Dislocations found in real materials are typically mixed, meaning that they have arranged into lattice planes (imagine stacking oranges in a grocery, each of the.

This causes a hardening of the metal as deformation progresses. In addition, adding pinning points that inhibit the motion of dislocations, such as alloying elements, can introduce stress fields that ultimately strengthen the material by requiring a higher applied stress to overcome the pinning stress and continue dislocation motion. The effects of strain hardening by accumulation of dislocations and the grain structure formed at high strain can be removed by appropriate heat treatment annealing which promotes the recovery and subsequent recrystallisation of the material.

The combined processing techniques of work hardening and annealing allow for control over dislocation density, the degree of dislocation entanglement, and ultimately the yield strength of the material. Climb[ edit ] Dislocations can slip in planes containing both the dislocation line and the Burgers vector. For a screw dislocation, the dislocation line and the Burgers vector are parallel, so the dislocation may slip in any plane containing the dislocation.

For an edge dislocation, the dislocation and the Burgers vector are perpendicular, so there is one plane in which the dislocation can slip. There is an alternative mechanism of dislocation motion, fundamentally different from slip, that allows an edge dislocation to move out of its slip plane, known as dislocation climb.

Dislocation climb allows an edge dislocation to move perpendicular to its slip plane. A creep mechanism involving only dislocation climb, also known as Harper-Dorn creepcan occur under certain conditions.

Furthermore, the 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. 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 bthe xy area of this triangle is. The triangle edge length, x, is equal to the length of a face diagonal, as indicated in Figure a.

In a manner similar to Figure 7. Now, using arrows, indicate two different slip directions within this plane. The arrows indicate two different type directions. You might find Figure 3. The arrows indicate three different type directions.

Also, since the magnitudes of these Burgers vectors may be determined from the following equation: You may want to consult Table 3. Its unit cell is shown in Figure 3. Also, simple cubic is the crystal structure for the edge dislocation of Figure 4.

You may also want to consult the answer to Concept Check 7.

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

This Concept Check asks that we select the slip system for simple cubic from four possibilities. Thus, the Burgers vector will lie in a -type direction. Also, the unit slip distance is a i. Determine the magnitude of the Schmid factor for an FCC single crystal oriented with its [] direction parallel to the loading axis. We are asked to compute the Schmid factor for an FCC crystal oriented with its [] direction parallel to the loading axis.

With this scheme, slip may occur on the plane and in the [1 1 0] direction as noted in the figure below. The angle between the [] and [1 1 0] directions, l, may be determined using Equation 7. Again from Equation 7.

If the critical resolved shear stress is 0. If not, what stress will be necessary? 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. Three possible slip directions make angles of 6.

We are asked to compute the critical resolved shear stress for Al. Cosines for the possible l values are given below. If slip occurs on a plane and in a [ 1 01] direction, and is initiated at an applied tensile stress of 1. This problem asks that we compute the critical resolved shear stress for silver. In order to do this, we must employ Equation 7. The angle l is the angle between the tensile axis i. The angle l may be determined using Equation 7. If the critical resolved shear stress for this material is 1.

In order to solve this problem it is necessary to employ Equation 7. For each of these three slip systems, the f will be the same i. The angle f may be determined using Equation 7. Again, using Equation 7. Now, we must determine the value of l for the [10 1 ] slip system that is, the angle between the [] and [10 1 ] directions. Again using Equation 7. If the magnitude of this stress is.

For each of these three slip systems, the l will be the same i. This angle l may be determined using Equation 7. 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.

Now, from the definition of engineering strain Equation 6. For one specimen, the initial and deformed radii are 16 mm and 11 mm, respectively. The second specimen, with an initial radius of 1 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation. In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such.

Their original and deformed dimensions are as follows: Circular diameter, mm Rectangular mm Original dimensions Deformed dimensions Which of these specimens will be the hardest after plastic deformation, and why? The hardest specimen will be the one that has experienced the greatest degree of cold work.

If its cold-worked radius is 10 mm 0. For a cylindrical specimen, Equation 7.

Finally, using Equation 6. For copper, the critical resolved shear stress is. If it is known that the value of A for copper is MPa-mm 0.

We are asked in this problem to compute the critical resolved shear stress at a dislocation density of 10 7 mm.

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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. Below is shown a square grid onto which is superimposed the recrystallized regions from the micrograph. 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. Time min Grain Diameter mm a What was the original grain diameter? You are asked to reduce its average grain diameter to mm.

## For each of edge, screw, and mixed dislocations, cite the

If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

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- Problem 7.4 Chapter 7

Yes, it is possible to reduce the average grain diameter of an undeformed alloy specimen from mm to mm. In order to do this, plastically deform the material at room temperature i. Estimate the yield strength of this alloy after it has been heated to C for s, if it is known that the value of k y is 1.

The tensile strength corresponding to a Brinell hardness of 5 may be determined using Equation 6. Therefore, it is possible to meet both of these criteria by plastically deforming the steel.

According to Figure 6. Furthermore, from Figure 7. Therefore, it is not possible to meet both of these criteria by plastically deforming brass.

D3 A cylindrical specimen of cold-worked steel has a Brinell hardness of If the metal may be cold worked, decide which of the following are candidates: For each of these alloys, the minimum cold work necessary to achieve the yield strength may be determined from Figure 7.

These data are tabulated below. D5 A cylindrical rod of steel originally Furthermore, the final diameter must be 10 mm 0. Explain how this may be accomplished. First let us calculate the percent cold work and attendant tensile strength and ductility if the drawing is carried out without interruption.

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.