A cylindrical rod of copper originally 16.0mm in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked yield strength in excess of 250 MPa and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 11.3 mm. Explain how this may be accomplished.
load: \( \sigma \) = 1.1 MPa in the [001] direction
slip system: (111)\( [ \bar{1} 11] \)
\( \tau _{crss} \) : critical resolved shear stress
\( \tau _{crss} \) = \( \sigma _{y}( cos\phi cos\lambda ) \)
\( \vec{U} \cdot \vec{V} = | \vec{U} | | \vec{V} | cos \theta \) where \( \theta \) is \( \angle \) between \( \vec{U} \) and \( \vec{V} \)
\( \lambda = cos^{-1} \left[ {(0)(-1) + (0)(0) + (1)(1) \over \sqrt{ [ (0)^2+(0)^2+(1)^2][(-1)^2+(0)^2+(1)^2]} } \right] \)
\( = cos^{-1}({1 \over \sqrt{2}}) = 45.0^\circ \)
\( \phi = cos^{-1} \left[ {(0)(1) + (0)(1) + (1)(1) \over \sqrt{ [ (0)^2+(0)^2+(1)^2][(1)^2+(1)^2+(1)^2]} } \right] \)
\( = cos^{-1}({1 \over \sqrt{3}}) = 54.7^\circ \)
\(\tau_{crss}\) = (1.1 MPa)[cos(54.7\( ^\circ \))cos(45.0\( ^\circ \))]
\(\tau_{crss}\) = 0.45 MPa