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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
Answer: All three slip systems ((110), (011), and (10−1)) are approximately equally favorably oriented for the given applied tensile stress in the [100] direction, with a resolved shear stress of about 2.31 MPa.

Step by step solution

01

Determine the angles between the stress direction and the slip directions

We'll find the angle between \([100]\) and \([1 \overline{1}1]\) using the formula for the angle between two vectors: \(\theta = cos^{-1}\frac{a \cdot b}{|a||b|}\), \[ \theta = cos^{-1}\frac{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} }{|\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}||\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}|}\] \[ \theta = cos^{-1}\frac{1}{\sqrt{1^2 + 0^2 + 0^2}\sqrt{1^2 + (-1)^2 + 1^2}} = cos^{-1}\frac{1}{\sqrt{3}} \] $$\phi = cos^{-1} \frac{1}{\sqrt{3}} \approx 54.7°$$
02

Determine the angles between the stress direction and slip plane normals

Now we need to determine the angles between the stress direction (\([100]\)) and each of the slip plane normals. For this, we'll use the given slip planes (\(\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{bmatrix}\)): $$\lambda_{(110)} = cos^{-1} \frac{1}{\sqrt{1^2 + 1^2}} = 45°$$ $$\lambda_{(011)} = cos^{-1} \frac{1}{\sqrt{1^2 + 1^2}} = 45°$$ $$\lambda_{(10 \overline{1})} = cos^{-1} \frac{1}{\sqrt{1^2 + 1^2}} = 45°$$
03

Calculate the resolved shear stress for each slip system

Now that we have the angles between the stress direction and the slip directions, as well as the angles between the stress direction and the slip plane normals, we can calculate the resolved shear stress for each slip system using the formula: \(\tau_{R} = \sigma*cos(\phi)*sin(\lambda)\) with \(\sigma = 4.0 \mathrm{MPa}\). \(\tau_{R_{(110)}} = (4.0 \mathrm{MPa}) * cos(54.7°) * sin(45°)\approx 2.31 \mathrm{MPa}\) \(\tau_{R_{(011)}} = (4.0 \mathrm{MPa}) * cos(54.7°) * sin(45°)\approx 2.31 \mathrm{MPa}\) \(\tau_{R_{(10 \overline{1})}} = (4.0 \mathrm{MPa}) * cos(54.7°) * sin(45°)\approx 2.31 \mathrm{MPa}\)
04

Determine the most favorably oriented slip system(s)

Since all three resolved shear stress values are approximately equal at \(2.31 \mathrm{MPa}\), this means that all three slip systems (\((110)\), \((011)\), and \((10 \overline{1})\)) are most favorably oriented for the given applied tensile stress in the \([100]\) direction.

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Most popular questions from this chapter

(a) What is the driving force for recrystallization? (b) For grain growth?

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

Equations 7.1a and 7.1b, expressions for Burgers vectors for \(\mathrm{FCC}\) and BCC crystal structures, are of the form $$ \mathbf{b}=\frac{a}{2}\langle u v w\rangle $$ where \(a\) is the unit cell edge length. Also, because the magnitudes of these Burgers vectors may be determined from the following equation: $$ |\mathbf{b}|=\frac{a}{2}\left(u^{2}+v^{2}+w^{2}\right)^{1 / 2} $$ determine values of \(|\mathbf{b}|\) for aluminum and chromium. You may want to consult Table 3.1.

One slip system for the BCC crystal structure is \(\\{110\\}(111\rangle\). In a manner similar to Figure \(7.6 b\), sketch a \\{110\\}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different \(\langle 111\rangle\) slip directions within this plane.

Consider a single crystal of silver oriented such that a tensile stress is applied along a \([001]\) direction. If slip occurs on a (111) plane and in a [ \(\overline{101}\) ] direction, and is initiated at an applied tensile stress of \(1.1 \mathrm{MPa}\) (160 psi), compute the critical resolved shear stress.
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