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Chapter 7: Problem 33

Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \(\tau_{\text {crss }}\) is a function of the dislocation density \(\rho_{D}\) as $$ \tau_{\mathrm{crss}}=\tau_{0}+A \sqrt{\rho_{D}} $$ where \(\tau_{0}\) and \(A\) are constants. For copper, the critical resolved shear stress is \(0.69 \mathrm{MPa}\) (100 psi) at a dislocation density of \(10^{4} \mathrm{~mm}^{-2}\). If it is known that the value of \(\tau_{0}\) for copper is \(0.069 \mathrm{MPa}\) (10 psi), compute \(\tau_{\text {crss }}\) at a dislocation density of \(10^{6} \mathrm{~mm}^{-2}\).

Short Answer

Expert verified
Question: Compute the critical resolved shear stress for copper at a dislocation density of \(10^6 \, \mathrm{mm}^{-2}\). Answer: The critical resolved shear stress for copper at a dislocation density of \(10^6 \, \mathrm{mm}^{-2}\) is approximately \(2.85 \, \mathrm{MPa}\).

Step by step solution

01

Calculate the value of constant A

Using the given equation \(\tau_{\mathrm{crss}}=\tau_{0}+A \sqrt{\rho_{D}}\), we can plug in the known values for copper and solve for the constant \(A\): Given values: - \(\tau_{\text{crss}} = 0.69\,\text{MPa}\) - \(\rho_{D}=10^{4}\,\text{mm}^{-2}\) - \(\tau_{0} = 0.069\,\text{MPa}\) Rearrange the equation to get: $$ A = \frac{\tau_{\text{crss}}-\tau_0}{\sqrt{\rho_D}} $$ Now, substitute the given values and solve for \(A\): $$ A = \frac{0.69\,\text{MPa} - 0.069\,\text{MPa}}{\sqrt{10^4\,\text{mm}^{-2}}} $$ Calculate the value of \(A\): $$A \approx 0.0621\,\text{MPa}\,\text{mm}$$
02

Compute new \(\tau_{\text{crss}}\) at different dislocation density

Now, we are given a new dislocation density \(\rho_{D} = 10^6\,\text{mm}^{-2}\), and we need to find the new value of \(\tau_{\text{crss}}\). Using the equation: $$ \tau_{\mathrm{crss}}=\tau_{0}+A \sqrt{\rho_{D}} $$ Substitute the values of \(\tau_{0}\), \(A\), and new \(\rho_{D}\): $$ \tau_{\text{crss}} = 0.069\,\text{MPa} + 0.0621\,\text{MPa}\,\text{mm} \times \sqrt{10^6\,\text{mm}^{-2}} $$ Calculate the new value of \(\tau_{\text{crss}}\): $$ \tau_{\text{crss}} \approx 2.85\,\text{MPa} $$ Thus, the new value of critical resolved shear stress \(\tau_{\text{crss}}\) for copper is \(\approx 2.85\, \mathrm{MPa}\) at a dislocation density of \(10^6\, \mathrm{mm}^{-2}\).

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Most popular questions from this chapter

The lower yield point for an iron that has an († average grain diameter of \(1 \times 10^{-2} \mathrm{~mm}\) is 230 MPa (33,000 psi). At a grain diameter of \(6 \times 10^{-3}\) \(\mathrm{mm}\), the yield point increases to \(275 \mathrm{MPa}(40,000\) psi). At what grain diameter will the lower yield point be \(310 \mathrm{MPa}(45,000 \mathrm{psi}) ?\)

For a brass alloy, the stress at which plastic deformation begins is \(345 \mathrm{MPa}(50,000 \mathrm{psi})\), and the modulus of elasticity is \(103 \mathrm{GPa}\left(15.0 \times 10^{6} \mathrm{psi}\right)\) (a) What is the maximum load that can be applied to a specimen with a cross- sectional area of \(130 \mathrm{~mm}^{2}\left(0.2 \mathrm{in} .{ }^{2}\right)\) without plastic deformation? (b) If the original specimen length is \(76 \mathrm{~mm}(3.0\) in.), what is the maximum length to which it can] be stretched without causing plastic deformation?

Consider a cylindrical nickel wire \(2.0 \mathrm{~mm}\) \(\left(0.08\right.\) in.) in diameter and \(3 \times 10^{4} \mathrm{~mm}\) (1200 in.) long. Calculate its elongation when a load of \(300 \mathrm{~N}\left(67 \mathrm{lb}_{t}\right)\) is applied. Assume that the defor-

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are \(30.00\) and \(30.04 \mathrm{~mm}\), respectively, and its final length is \(105.20 \mathrm{~mm}\), compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are \(65.5\) and \(25.4\) GPa, respectively.

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.
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