Question

(10 points) Consider a single crystal of some hypothetical metal that has the FCC crystal structure, and its critical resolved shear stress is 3.42 MPa. The crystal is oriented such that a tensile stress is applied at an angle of 39.2 degrees to the slip plane. Slip can occur in two directions( 18.4 and 74.2 degrees to the tensile force). (a) (5 points) Which direction, is slip favored

Answer 1

Complete Question

(10 points) Consider a single crystal of some hypothetical metal that has the FCC crystal structure, and its critical resolved shear stress is 3.42 MPa. The crystal is oriented such that a tensile stress is applied at an angle of 39.2 degrees to the slip plane. Slip can occur in two directions( 18.4 and 74.2 degrees to the tensile force).

(a) (5 points) Which direction, is slip favored?

b)  (5 points)  What is the yield strength of this crystal during this tensile test ?

Answer:

a

the slip would occur 18.4° to the tensile force

b

The yield strength is    $\sigma_y = 4.65\ MPa$

Explanation:

From the question we are told that

    The Resolved shear force is $\sigma = 3.2MPa$

     The angle in which the tensile stress is applied is $\O = 39.2^o$

      The first direction of slip is $\theta _ 1 =18.4^o$

      The second direction of the slip is $\theta_2 = 74.2^o$

Generally the condition for direction in which slip is likely to occur(

direction in which slip is favored )   is that  $(cos( \O) cos (\theta) )$

Must be Maximum for that direction

   Since $Cos (\O)$ is constant for both direction we would  look at the the cos of the the angle for both direction

      $Cos (\theta_ 1) = Cos(18.4^o) =0.9488$

      $Cos (\theta_ 2) = Cos(74.2^o) =0.2722$

From this calculation we can see that the slip would occur 18.4° to the tensile force

   Generally critical resolved shear stress is mathematically represented as

           $\sigma = \sigma_y * (cos(\O) cos(\theta_1))$

Where $\sigma_y$ is the  yield strength

Making  $\sigma_y$ the subject

       $\sigma_y = \frac{\sigma }{[cos (\O) cos(\theta_1)]}$

Substituting value

             $\sigma_y = \frac{3.4*10^{6}}{cos (39.2) (cos 18.4)}$

                 $\sigma_y = 4.65\ MPa$