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 29.8 MPa, what applied stress (in MPa) will be necessary to cause the single crystal to yield?

The applied stress required to cause the single crystal to yield is 60.96MPa

Explanation:

Given:

Ф = 42.7°

λ = 48.3°

Critical resolved shear stress = 29.8 MPa

Applied stress = ?

We know,

Critical resolved shear stress = (applied stress) ( cos 42.7°) ( cos 48.3°)

29.8 MPa = s X (0.735) X (0.665)

Therefore, the applied stress required to cause the single crystal to yield is 60.96MPa