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Chapter 7: Problem 16

(a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the \([100]\) direction. If the magnitude of this stress is \(4.0 \mathrm{MPa}\), compute the resolved shear stress in the \([1 \overline{11}]\) direction on each of the \((110),(011)\), and \((10 \overline{1})\) planes. (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented?

Short Answer

Expert verified
Consider the slip systems with a slip direction of [1 -1 1] on planes (110), (011), and (10 -1). Answer: The most favorably oriented slip system(s) for this situation are the (110) and (10 -1) slip systems, with a resolved shear stress of (2√2)/(√3) MPa each.

Step by step solution

01

Calculate the angle between slip direction and stress direction vector (\(\alpha\)) for all slip systems

For all the slip systems, the slip direction is the \([1, \overline{1}, 1]\). First, we'll find the angle between this slip direction and the stress direction, \([100]\). \(\cos{\alpha} = \frac{\vec{s} \cdot \vec{t} }{|\vec{s}| |\vec{t}|}\) Where \(\vec{s}\) represents the slip direction vector and \(\vec{t}\) represents the stress direction vector \(\cos{\alpha} = \frac{[1, \overline{1}, 1] \cdot [1, 0, 0]}{|[1, \overline{1}, 1]| |[1, 0, 0]|} = \frac{1}{\sqrt{1^2 + (-1)^2 + 1^2}\sqrt{1^2 + 0^2 + 0^2}} = \frac{1}{\sqrt{6}}\)
02

Calculate the angle between the normal slip plane and stress direction vector (\(\beta\)) for all slip systems

We need to find out \(\beta\) for each slip system, where \(\beta\) is the angle between the normal slip plane and the stress direction vector, \([100]\). a. For the \((110)\) plane, the plane vector is \(\vec{p} = [1, 1, 0]\). So, \(\cos{\beta} = \frac{[1, 1, 0] \cdot [1, 0, 0]}{|[1, 1, 0]| |[1, 0, 0]|} = \frac{1}{\sqrt{2}}\) b. For the \((011)\) plane, the plane vector is \(\vec{p} = [0, 1, 1]\). So, \(\cos{\beta} = \frac{[0, 1, 1] \cdot [1, 0, 0]}{|[0, 1, 1]| |[1, 0, 0]|} = 0\) c. For the \((10\overline{1})\) plane, the plane vector is \(\vec{p} = [1, 0, -1]\). So, \(\cos{\beta} = \frac{[1, 0, -1] \cdot [1, 0, 0]}{|[1, 0, -1]| |[1, 0, 0]|} = \frac{1}{\sqrt{2}}\)
03

Calculate the resolved shear stress for all slip systems using Schmid's law

Now we'll plug the values of \(\cos{\alpha}\) and \(\cos{\beta}\) into Schmid's law for each slip system and find the resolved shear stress. a. For the \((110)\) slip system: \(\tau_{res} = \sigma \cos{\alpha} \cos{\beta} = 4.0 \cdot \frac{1}{\sqrt{6}} \cdot \frac{1}{\sqrt{2}} = \frac{4\sqrt{2}}{\sqrt{12}} = \frac{4\sqrt{2}}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}} \mathrm{MPa}\) b. For the \((011)\) slip system: \(\tau_{res} = \sigma \cos{\alpha} \cos{\beta} = 4.0 \cdot \frac{1}{\sqrt{6}} \cdot 0 = 0 \mathrm{MPa}\) c. For the \((10\overline{1})\) slip system: \(\tau_{res} = \sigma \cos{\alpha} \cos{\beta} = 4.0 \cdot \frac{1}{\sqrt{6}} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{3}} \mathrm{MPa}\)
04

Identify the most favorably oriented slip system(s)

Now we'll compare the resolved shear stress values we found for each slip system: - \((110)\) slip system: \(\tau_{res} = \frac{2\sqrt{2}}{\sqrt{3}} \mathrm{MPa}\) - \((011)\) slip system: \(\tau_{res} = 0 \mathrm{MPa}\) - \((10\overline{1})\) slip system: \(\tau_{res} = \frac{2\sqrt{2}}{\sqrt{3}} \mathrm{MPa}\) The most favorably oriented slip system(s) are the ones with the highest resolved shear stress value. In this case, the slip systems \((110)\) and \((10\overline{1})\) have the highest resolved shear stress values, which are equal to \(\frac{2\sqrt{2}}{\sqrt{3}} \mathrm{MPa}\).

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Most popular questions from this chapter

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30\times10^{6 psi) and an } original diameter of \(10.2 \mathrm{~mm}(0.40\) in.) experiences only elastic deformation when a tensile load of \(8900 \mathrm{~N}\left(2000 \mathrm{lb}_{i}\right)\) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is \(0.25 \mathrm{~mm}\) \((0.010\) in.).

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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}

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A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is \(0.5 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [101], [10\overline{1} ] \text { , and } [ 0 \overline { 1 1 } ] \text { } directions.
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