Question

Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, respectively, with the tensile axis. If the critical resolved shear stress is 29.4 MPa, what applied stress (in MPa) will be necessary to cause the single crystal to yield?

Answer 1

Given:

slip direction angle,
`\phi = 42.7^(\circ)`

angle normal to slip plne,
`\lambda = 48.3^(\circ)`

critical resolved shear stress,
`\tau_(r) = 29.4 MPa`

Solution:

To calculate the applied stress,
`\sigma_(a)` the formula for critical resolved shear stress is given by:

`\tau_(r) = \sigma_(a) cos\lambda cos\phi` (1)

Putting the given values in eqn (1):

`29.4* 10^(6) = \sigma_(a) * cos48.3^(\circ)* cos 42.7^(\circ)`

`\sigma _(a) = 60.14 MPa`

Therefore, necessary applied stress is 60.14 MPa