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Chapter 6: Problem 18

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.

Short Answer

Expert verified
Answer: The critical resolved shear stress is approximately 4.197 MPa.

Step by step solution

01

Calculate the dot product of the stress axis and slip direction

First, we must find the angle between the stress axis (direction [112]) and the slip direction (direction [011]). To do this, we will use the dot product, \(\cos{\theta_1} = \frac{\text{dot product}}{\text{product of magnitudes}}\). Dot product of the stress axis and the slip direction: (1)(0) + (1)(1) + (2)(1) = 1 + 2 = 3 Product of magnitudes of the stress axis and slip direction: \(|[112]| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}\) \(|[011]| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2}\) Hence, \(\cos{\theta_1} = \frac{3}{\sqrt{6} \cdot \sqrt{2}} = \frac{3}{2\sqrt{3}}\)
02

Calculate the dot product of the stress axis and slip plane normal

Next, we need to find the angle between the stress axis (direction [112]) and the slip plane normal (direction [111] as it is normal to the (111) plane). We will again use the dot product formula. Dot product of the stress axis and the slip plane normal: (1)(1) + (1)(1) + (2)(1) = 1 + 1 + 2 = 4 Product of magnitudes of the stress axis and slip plane normal: \(|[112]| = \sqrt{6}\) (already calculated) \(|[111]| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}\) Hence, \(\cos{\theta_2} = \frac{4}{\sqrt{6} \cdot \sqrt{3}} = \frac{4}{3\sqrt{2}}\)
03

Calculate the critical resolved shear stress using Schmid's Law

Now, we will use the Schmid's Law formula, which states that the critical resolved shear stress is equal to the applied stress times \(\cos{\theta_1}\) times \(\cos{\theta_2}\). Critical Resolved Shear Stress = Applied Stress × \(\cos{\theta_1}\) × \(\cos{\theta_2}\) \(= 5.12 \, \mathrm{MPa} \times \left(\frac{3}{2\sqrt{3}}\right) \times \left(\frac{4}{3\sqrt{2}}\right)\) \( = 5.12 \, \mathrm{MPa} \times \frac{3 \cdot 4}{(2\sqrt{3})(3\sqrt{2})}\) \( = 5.12 \, \mathrm{MPa} \times \frac{12}{(6\sqrt{6})}\) \( = 5.12 \, \mathrm{MPa} \times \frac{2}{\sqrt{6}}\) The critical resolved shear stress is approximately: \( = 4.197 \, \mathrm{MPa}\).

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Most popular questions from this chapter

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.

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

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 \(6.25\) 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, $$ E \propto\left(\frac{d F}{d r}\right)_{r_{0}} $$ 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 \(2.8 b\) ), 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 \(d F / d r\).

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation, the circular cross section is to remain circular, and the rectangular is to remain rectangular. Their original and deformed dimensions are as follows: \begin{tabular}{lcc} \hline & Circular \((\) diameter, \(\boldsymbol{m m})\) & Rectangular (mm) \\ \hline Original dimensions & \(18.0\) & \(20 \times 50\) \\ \hline Deformed dimensions & \multicolumn{1}{c}{\(15.9\)} & \(13.7 \times 55.1\) \\\ \hline \end{tabular} Which of these specimens will be the hardest after plastic deformation, and why?

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}
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