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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

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?

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?

In Section \(2.6\), it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.31\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is,Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system), using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus- \(r\) curve (Figure 2.10b), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\) - 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of \(5.12 \mathrm{MPa}\), compute the critical resolved shear stress.

A cylindrical specimen of stainless steel having a diameter of \(12.8 \mathrm{~mm}(0.505 \mathrm{in}\).) and a gauge length of \(50.800 \mathrm{~mm}(2.000 \mathrm{in}\) ) is pulled in tension. Use he load-elongation characteristics shown in the ollowing table to complete parts (a) through (f). (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of \(0.002\). (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience.
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