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Chapter 6: Problem 17

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(2.4 \mathrm{MPa}\).

Short Answer

Expert verified
Answer: The crystal yields when the applied stress along the [121] direction is approximately 10.3 MPa.

Step by step solution

01

Calculate the angle between the stress direction and the slip direction

We can use the following formula to calculate the angle between the stress direction [uvw] and the slip direction [lmn], where u, v, w, l, m, and n are the respective indices: $$\cos \alpha = \frac {u \cdot l + v \cdot m + w \cdot n} {\sqrt {(u^2+v^2+w^2)(l^2+m^2+n^2)}}$$ From the problem, we have the stress direction [121], and the slip direction [111]. Substituting the values in the above equation, we get: $$\cos \alpha = \frac {1 \cdot 1 + 2 \cdot 1 + 1 \cdot 1} {\sqrt {(1^2+2^2+1^2)(1^2+1^2+1^2)}}$$
02

Compute the angle α

After simplifying the expression in Step 1, we can now compute the angle α: $$\cos \alpha = \frac {1+2+1} {\sqrt {(6)(3)}} = \frac{4}{3\sqrt{2}}$$ Now, find the angle α: $$\alpha = \arccos{\frac{4}{3\sqrt{2}}}$$
03

Calculate the angle between the slip plane and stress plane

We can use the following formula to calculate the angle between the slip plane (hkl) and the stress plane [uvw], where h, k, l, u, v, and w are the respective indices: $$\cos \phi = \frac {h \cdot u + k \cdot v + l \cdot w} {\sqrt {(h^2+k^2+l^2)(u^2+v^2+w^2)}}$$ From the problem, we have the slip plane (101) and the stress direction [121]. Substituting the values in the above equation, we get: $$\cos \phi = \frac {1 \cdot 1 + 0 + 1 \cdot 1} {\sqrt {(1^2+0^2+1^2)(1^2+2^2+1^2)}}$$
04

Compute the angle φ

After simplifying the expression in Step 3, we can now compute the angle φ: $$\cos \phi = \frac {1+0+1} {\sqrt {(2)(6)}} = \frac{2}{2\sqrt{3}}$$ Now, find the angle φ: $$\phi = \arccos{\frac{2}{2\sqrt{3}}}$$
05

Calculate the critical resolved shear stress

We can now determine the stress at which the crystal yields using the equation for critical resolved shear stress: $$\text{Critical Resolved Shear Stress} = \frac {\text{Yielding Stress} \cdot \cos \alpha \cdot \cos \phi} {2}$$ We can rearrange this equation to solve for the Yielding Stress: $$\text{Yielding Stress} = \frac{2 \cdot \text{Critical Resolved Shear Stress}}{\cos \alpha \cdot \cos \phi}$$ Now, we are given that the critical resolved shear stress is \(2.4 \mathrm{MPa}\). Substituting this value and the angles α and φ into the equation, we get: $$\text{Yielding Stress} = \frac{2 \cdot 2.4 \mathrm{MPa}}{\frac{4}{3\sqrt{2}} \cdot \frac{2}{2\sqrt{3}}}$$
06

Compute the Yielding Stress

Finally, compute the Yielding Stress: $$\text{Yielding Stress} \approx 10.3 \mathrm{MPa}$$ So, the crystal yields when the applied stress along the [121] direction is approximately 10.3 MPa.

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Most popular questions from this chapter

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

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.

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{tabular}{lcc} \hline Grain diameter ( mm) & Yield Strength (MPa) & Heat Treating Time (h) \\\ \hline \(0.028\) & 300 & 10 \\ \hline \(0.010\) & 385 & 1 \\ \hline \end{tabular}

Briefly explain why HCP metals are typically more brittle than \(\mathrm{FCC}\) and \(\mathrm{BCC}\) metals.

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.
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