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 direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer 1

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 $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

  The critical yield resolved shear stress is  $\sigma = 2.9Mpa$

First we obtain the angle  $\lambda$ between the slip direction [121] and  [111]

             $\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

             $\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

                $= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

                 $= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

              $\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

 Substituting 1 for $u_1$ , 2 for  $v_1$ , 1 for $w_1$ , 1 for  $u_2$, 0 for  $v_2$, and 1 for  $w_2$

            $\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

               $\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

                   $= 54.74 ^o$

   The stress is mathematically represented as

              $\sigma = \frac{\tau_c}{cos \O cos \lambda }$

                  $= \frac{2.9}{cos 54.74^o cos 61.87^o}$

                  $= \frac{2.9}{0.2722}$

             $\sigma = 10. 655 MPa$