Answer:
the magnitude of the stress necessary to cause slip to occur on the (111) plane in the [tex][1 0 \overline 1][/tex] direction is 4.85 MPa
Explanation:
From the given information;
To determine the angle [tex]\phi[/tex] between the direction [100] and [111]; we have:
[tex]\phi = cos ^{-1} [\dfrac{u_1u_2+v_1v_2+w_1w_2}{ \sqrt{(u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) }}}[/tex]
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 :
[tex]\phi = cos ^{-1} [\dfrac{1*1+0*1+0*1}{ \sqrt{(1^2+0^2+0^2) (1^2+1^2+1^2) }}}[/tex]
[tex]\phi = 54.7^0[/tex]
To determine the angle [tex]\lambda[/tex] for the slip direction [tex][1 0 \overline 1][/tex]
[tex]\lambda= cos ^{-1} [\dfrac{u_1u_2+v_1v_2+w_1w_2}{ \sqrt{(u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) }}}[/tex]
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 :
[tex]\lambda = cos ^{-1} [\dfrac{1*1+0*0+(0*-1)}{ \sqrt{(1^2+0^2+0^2) (1^2+0^2+(-1^2)) }}}[/tex]
[tex]\phi = cos ^{-1} [\dfrac{1}{\sqrt{2}}][/tex]
[tex]\lambda = 45^0[/tex]
The yield strength for the slip process [tex][1 0 \overline 1][/tex] can now be calculated as:
[tex]\sigma_x = \dfrac{t_{xr}}{cos \phi \ \ cos \lambda}[/tex]
where
[tex]t_{xr}[/tex] = 1.98 MPa
[tex]\sigma_x = \dfrac{1.98}{cos 54.7^0 \ \ cos 45}[/tex]
[tex]\sigma_x = \dfrac{1.98}{0.5779 *0.7071}[/tex]
[tex]\mathbf{\sigma _x = 4.85 \ MPa}[/tex]
Hence, the magnitude of the stress necessary to cause slip to occur on the (111) plane in the [tex][1 0 \overline 1][/tex] direction is 4.85 MPa
