Complete Question
Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.
Answer:
The stress is [tex]\sigma = 10. 655 MPa[/tex]
Explanation:
From the question we are told that
The critical yield resolved shear stress is [tex]\sigma = 2.9Mpa[/tex]
First we obtain the angle [tex]\lambda[/tex] between the slip direction [121] and [111]
[tex]\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ][/tex]
Where [tex]u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2[/tex] are the directional indices
[tex]\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ][/tex]
[tex]= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ][/tex]
[tex]= 61.87^0[/tex]
Next is to obtain the angle [tex]\O[/tex] between the direction [121] and [101]
[tex]\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ][/tex]
Substituting 1 for [tex]u_1[/tex] , 2 for [tex]v_1[/tex] , 1 for [tex]w_1[/tex] , 1 for [tex]u_2[/tex], 0 for [tex]v_2[/tex], and 1 for [tex]w_2[/tex]
[tex]\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ][/tex]
[tex]\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ][/tex]
[tex]= 54.74 ^o[/tex]
The stress is mathematically represented as
[tex]\sigma = \frac{\tau_c}{cos \O cos \lambda }[/tex]
[tex]= \frac{2.9}{cos 54.74^o cos 61.87^o}[/tex]
[tex]= \frac{2.9}{0.2722}[/tex]
[tex]\sigma = 10. 655 MPa[/tex]
