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Chapter 7: Problem 18
The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.
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A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of \(28.1^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(62.4^{\circ}, 72.0^{\circ}\), and \(81.1^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(1.95 \mathrm{MPa}(280 \mathrm{psi})\), determine the critical resolved shear stress for aluminum.
(a) What is the approximate ductility (\%EL) of a brass that has a yield strength of \(275 \mathrm{MPa}\) \((40,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 690 \(\mathrm{MPa}(100,000 \mathrm{psi})\) ?
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.
Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.
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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