Answer to Question #141918 in General Chemistry for Angel

Question

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1° and 47.9°, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa, will an applied stress of 45 MPa cause the single crystal to yield? If not, what stress will be necessary?

Answer 1

From the data given;

$\phi=43.1°,\lambda=47.9°C$

Values of the critical resolved sheer stress and applied tensile stress are $20.7MPa$ and $45MPa$

From the equation in the Schmidt factor, where

$(\tau)=FCos \lambda Cos \psi$

$\tau_{R}=\sigma Cos \phi Cos \lambda$

$=(45MPa)(Cos 43.1°C)( Cos 47.9°C)$ $=22.0MPa$

Since the resolved sheer stress$(22MPa)$ is greater than the critical resolved sheer stress $(20.7MPa)$, the single Crystal will yield.