Question

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 1.98 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.

Answer 1

the magnitude of the stress necessary to cause slip to occur on the (111) plane in the
`[1 0 \overline 1]` direction is 4.85 MPa

Step-by-step explanation:

From the given information;

To determine the angle
`\phi` between the direction [100] and [111]; we have:

`\phi = cos ^(-1) [(u_1u_2+v_1v_2+w_1w_2)/( √((u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) ))}`

where;

[u₁,v₁,w₁] and [u₂, v₂, w₂] are directional indices.

replacing 1 for u₁ , 0 for v₁ and 0 for w₁ ;

also replacing 1 for u₂, 1 for v₂ and 1 for w₂ ; we have :

`\phi = cos ^(-1) [(1*1+0*1+0*1)/( √((1^2+0^2+0^2) (1^2+1^2+1^2) ))}`

`\phi = 54.7^0`

To determine the angle
`\lambda` for the slip direction
`[1 0 \overline 1]`

`\lambda= cos ^(-1) [(u_1u_2+v_1v_2+w_1w_2)/( √((u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) ))}`

replacing 1 for u₁ , 0 for v₁ and 0 for w₁ ;

also replacing 1 for u₂, 1 for v₂ and -1 for w₂ ; we have :

`\lambda = cos ^(-1) [(1*1+0*0+(0*-1))/( √((1^2+0^2+0^2) (1^2+0^2+(-1^2)) ))}`

`\phi = cos ^(-1) [(1)/(√(2))]`

`\lambda = 45^0`

The yield strength for the slip process
`[1 0 \overline 1]` can now be calculated as:

`\sigma_x = (t_(xr))/(cos \phi \ \ cos \lambda)`

where

`t_(xr)` = 1.98 MPa

`\sigma_x = (1.98)/(cos 54.7^0 \ \ cos 45)`

`\sigma_x = (1.98)/(0.5779 *0.7071)`

`\mathbf{\sigma _x = 4.85 \ MPa}`

Hence, the magnitude of the stress necessary to cause slip to occur on the (111) plane in the
`[1 0 \overline 1]` direction is 4.85 MPa