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Chapter 7: Problem 3

Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.

Short Answer

Expert verified
Answer: No, two screw dislocations of opposite sign cannot fully annihilate each other like edge dislocations. When they interact, they create a new screw dislocation with a Burgers vector equal to the sum of the initial dislocations, but they do not return the lattice to a perfect state.

Step by step solution

01

Define Screw Dislocations

In crystallography, a screw dislocation is a type of linear defect that occurs within a crystal lattice. It is the result of the lattice atoms shifting along a helical path around the dislocation line. A screw dislocation can either be right-handed (positive) or left-handed (negative), depending on the direction of the helical atomic path around the dislocation.
02

Define Edge Dislocations

Edge dislocations, on the other hand, are linear defects that result from the termination of a plane of atoms within a crystal lattice. These dislocations cause an extra half-plane of atoms in the lattice and can be positive and negative as well.
03

Annihilation of Dislocations

Dislocations of opposite sign can annihilate each other to reduce the net Burgers vector, which describes the magnitude and direction of a dislocation. For example, when an edge dislocation with a positive Burgers vector meets a negative edge dislocation of the same magnitude, both lattice defects are canceled out, returning the lattice to a more perfect state.
04

Interaction of Screw Dislocations

Now let's consider the interaction between two screw dislocations with opposite signs. If they are brought close to each other, the helical atomic paths of the lattice atoms around both dislocation lines will not interlock or mean the perfect crystal lattice state. Instead, they form a new screw dislocation with a Burgers vector equal to the sum of two original dislocations. Since the right-handed and left-handed screw dislocations are of opposite nature, their Burgers vectors are non-collinear in the screw dislocation case.
05

Conclusion

In conclusion, it is not possible for two screw dislocations of opposite sign to fully annihilate each other like edge dislocations. When they interact, they create a new screw dislocation with a Burgers vector equal to the sum of the initial dislocations, but they do not return the lattice to a perfect state.

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Most popular questions from this chapter

As noted in Section \(3.15\), for single crystals of some substances, the physical properties are anisotropic that is, they depend on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general \([u v w]\) direction, \(E_{\text {uww }}\) is described by the relationship $$ \begin{aligned} \frac{1}{E_{u v w}}=& \frac{1}{E_{(100)}}-3\left(\frac{1}{E_{(100)}}-\frac{1}{E_{\\{111)}}\right) \\\ &\left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}\right) \end{aligned} $$ where \(E_{\\{100)}\) and \(E_{\langle 111\rangle}\) are the moduli of elasticity in the \([100]\) and \([111]\) directions, respectively; \(\alpha, \beta\), and \(\gamma\) are the cosines of the angles between \([u v w]\) and the respective [100], [010], and [001] directions. Verify that the \(E_{\\{110\rangle}\) values for aluminum, copper, and iron in Table \(3.4\) are correct.

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.

A cylindrical specimen of steel having a diameter of \(15.2 \mathrm{~mm}(0.60\) in.) and length of 250 \(\mathrm{mm}(10.0 \mathrm{in} .)\) is deformed elastically in tension with a force of \(48,900 \mathrm{~N}\left(11,000 \mathrm{lb}_{e}\right)\). Using the data contained in Table \(6.1\), determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease?

An aluminum bar \(125 \mathrm{~mm}(5.0\) in.) long and having a square cross section \(16.5 \mathrm{~mm}(0.65 \mathrm{in} .)\) on an edge is pulled in tension with a load of 66,700 \(\mathrm{N}\left(15,000 \mathrm{lb}_{f}\right)\) and experiences an elongation of \(0.43 \mathrm{~mm}\left(1.7 \times 10^{-2}\right.\) in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

The following yield strength, grain diameter, and heat treatment time (for grain growth) data were gathered for an iron specimen that was heat treated at \(800^{\circ} \mathrm{C}\). Using these data, compute the yield strength of a specimen that was heated at \(800^{\circ} \mathrm{C}\) for \(3 \mathrm{~h}\). Assume a value of 2 for \(n\), the grain diameter exponent. $$ \begin{array}{lcc} \hline \begin{array}{l} \text { Grain } \\ \text { diameter } \\ (\mathbf{m m}) \end{array} & \begin{array}{c} \text { Yield } \\ \text { Strength } \\ \text { (MPa) } \end{array} & \begin{array}{c} \text { Heat } \\ \text { Treating } \\ \text { Time (h) } \end{array} \\ \hline 0.028 & 300 & 10 \\ \hline 0.010 & 385 & 1 \\ \hline \end{array} $$
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