Question

Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 8 MPa, will the applied stress of 12 MPa cause the single crystal to yield? If not, what stress will be necessary?

Answer 1

19.5324 MPa

Step-by-step explanation:

Information provided

Angle between the normal to the slip plane with tensile axis,
`\alpha=60^(o)`

Angle by slip direction with tensile axis,
`\beta=35^(o)`

Critical resolved shear stress,
`\tau_(c)=8 MPa`

Applied stress
`\sigma=12 MPa`

Shear stress at slip plane

`\tau=\sigma cos\alpha cos\beta`

`\tau=12cos60^(o)cos35^(o)=4.915 MPa`

`\tau <\tau_(c)` hence crystal won’t yield

Applied stress,
`\sigma` for crystal to yield is given by

`\sigma=\frac {\tau_(c)}{cos\alpha cos\beta}`

`\sigma=\frac {8}{cos60cos35}=19.53239342 MPa`

`\sigma=19.5324 MPa`