List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

Short Answer

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Answer: The four major differences are: 1. Mechanism - Deformation by twinning involves changes in crystal lattice orientation through the formation of mirror-image regions (twins), while deformation by slip involves the movement of dislocations and shearing of the crystal structure. 2. Conditions of occurrence - Twinning usually occurs at low temperatures and high strain rates, whereas slip occurs at high temperatures and low strain rates. 3. Final result - Twinning results in a more intricate microstructure, often exhibiting better mechanical properties, while slip results in a permanent distortion of the crystal lattice with the formation of dislocations. 4. Ductility and strain hardening - Twinning is more common in materials with low ductility, while slip is more common in ductile materials. Strain hardening is more prominent in materials deformed by slip, whereas twinning causes less strain hardening due to twin boundaries acting as barriers to dislocation motion.

Step by step solution

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1. Mechanism

In deformation by twinning, the crystal lattice undergoes a change in its orientation through the formation of mirror-image regions (twins). This occurs across a specific crystallographic plane called the twinning plane. In contrast, deformation by slip involves the movement of dislocations through the crystal lattice, resulting in the shearing of the structure. This typically happens along the most densely packed atomic planes, like the close-packed planes.
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2. Conditions of occurrence

Deformation by twinning usually occurs at low temperatures and high strain rates, while deformation by slip mostly occurs at high temperatures and low strain rates. High strain rates increase the possibility of twinning rather than slip, as more energy is stored in the material during the deformation process. Similarly, deformation by slip is thermally activated, meaning it is more favorable at higher temperatures as atomic motion becomes easier.
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3. Final result

The final result of deformation by twinning is a more intricate and complex microstructure, with twin boundaries being difficult-to-move obstacles to further deformation. As a result, twinned materials often exhibit better mechanical properties, such as strength and toughness, compared to materials deformed by slip. In contrast, the final result of deformation by slip is a permanent distortion of the crystal lattice with the formation of dislocations.
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4. Ductility and strain hardening

Deformation by twinning usually happens in materials with low ductility, whereas deformation by slip is more common in ductile materials. Strain hardening (hardening of the material with increasing plastic deformation) tends to occur more prominently in materials deformed by slip due to the accumulation and interaction of dislocations. Twinning, on the other hand, causes less strain hardening as the twin boundaries can act as barriers to dislocation motion, limiting dislocation density buildup.

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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.

6.24 A cylindrical rod \(120 \mathrm{~mm}\) long and having a (- diameter of \(15.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(35,000 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(1.2 \times 10^{-2} \mathrm{~mm}\). Of the following materials listed, which are possible candidates? Justify your choice(s). \begin{tabular}{lccc} \hline & Modulus of Material & \begin{tabular}{c} \mathrm{ Yield } \(\\\ {\text { Elasticity }} \\ {\text { (GPa) }}\) & Strength (MPa) & Poisson's Ratio \\ \hline Aluminum alloy & 70 & 250 & \(0.33\) \\ \hline Titanium alloy & 105 & 850 & \(0.36\) \\ \hline Steel alloy & 205 & 550 & \(0.27\) \\ \hline Magnesium alloy & 45 & 170 & \(0.35\) \\ \hline \end{tabular} \end{tabular}

A specimen of copper having a rectangular cross section \(15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}(0.60 \mathrm{in} . \times 0.75 \mathrm{in} .)\) is pulled in tension with \(44,500 \mathrm{~N}\left(10,000 \mathrm{lb}_{\mathrm{f}}\right)\) force, producing only elastic deformation. Calculate the resulting strain.

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of \(60^{\circ}\) and \(35^{\circ}\), respectively, with the tensile axis. If the critical resolved shear stress is \(6.2 \mathrm{MPa}\) (900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary?

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of \(10.0\) \(\mathrm{mm}(0.39 \mathrm{in} .) .\) A tensile force of \(1500 \mathrm{~N}\left(340 \mathrm{lb}_{\mathrm{f}}\right)\) produces an elastic reduction in diameter of \(6.7 \times 10^{-4} \mathrm{~mm}\left(2.64 \times 10^{-5}\right.\) in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is \(0.35\).

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