Question

pts) ectively, with the tensile stress of 12 MPa (1750 psi) cause the single cry och Aangermal to slip plan and the slip direction are at angles of 60° axis. If the critical resolved shear stress is 6.2 l to yie that the normal to the slip plane Consider a metal single crystal oriented saod 35 res(go ps). nt, what stress nted such shear stress is 6.2 MPa (900 psi), will an applied If not, what stress and 35°, resp will be necessary? Schmid's Law TR OCoSod stress and slip direction,ge applied TR Ơcosфс0s Asangle between applied an angle between applied stress and normal to slip plane

Answer 1

Given: Angle between applied stress and normal to slip plane, $\phi$=60⁰

Angle between applied stress and slip direction, $\lambda$=35⁰

Critical resolved shear stress, $\tau_{crss}$=6.2 MPa

Applied stress, $\sigma$=12 MPa

The resolved shear stress using the expression

$\tau_{R}=\sigma*cos \phi*cos \lambda$

where $\tau_{R}$ is the Resolved shear stress

$\sigma$ is the Applied stress

$\phi$ is the Angle between applied stress and normal to slip plane

$\lambda$ is the Angle between applied stress and slip direction

TR 12 MPacos(60°) * cos (35°)

$\boldsymbol{\tau_{R}=4.91\ MPa}$

The single crystal will not yield as the resolved shear stress (4.91 MPa) is less than the critical resolved shear stress (6.2 MPa)

The magnitude of applied stress at which yielding occurs is obtained using the expression

CT88

where $\sigma_{y}$ is the Magnitude of applied stress at which yielding occurs or the yield strength

$\tau_{crss}$ is the Critical resolved shear stress

$\Rightarrow \sigma_{y}=\frac{6.2\ MPa}{cos(60^{\circ})*cos(35^{\circ})}$

$\Rightarrow \sigma_{y}=15.137\ MPa$

$\mathbf{\boldsymbol{\sigma}_{y}=15.14\ MPa}$

Thus the yielding will occur at an applied stress of $\boldsymbol{\sigma}$ _y=15.14 MPa