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

Answer 1

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

Step-by-step 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)

`s = (29.8)/(0.735 X 0.665) \\\\s = 60.96MPa`

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