Respuesta :
Answer:
Thus the most favorable slip option is for [1 1 0] plane.
Solution:
As per the question:
Magnitude of the stress, [tex]\sigma = 2.75\ MPa[/tex]
No. of atoms in Body Centered Cubic Crystal, n = 2
Now,
Resolved shear stress is given by:
[tex]\sigma_{R} = \sigma cos\theta cos\lambda [/tex]
where
[tex]\theta [/tex] = angle between slip plane and the tensile axis
[tex]\lambda [/tex] = angle between tensile axis and the slip direction
Direction of applied force: [0 1 0]
Plane: [1 1 0]
Direction: [tex][\bar{1}\ 1\ 1][/tex]
Also, we know that:
[tex]cos\theta = \frac{a_{1}a_{2} + b_{1}b_{2} +c_{1}c_{2}}{|a||b|}[/tex]
Thus
[tex]cos\theta = \frac{(0\times 1) + (1\times 1) + (0\times 0)}{1\times \sqrt{1^{2} + 1^{2}} + {0}^{2}} = \frac{1}{\sqrt{2}}[/tex]
[tex]cos\lambda = \frac{(0\times - 1) + (1\times 1) + (0\times 1)}{1\times \sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}[/tex]
[tex]\sigma_{R} = \sigma cos\theta cos\lambda = 2.75\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{3}}[/tex]
[tex]\sigma_{R} = 1.123\ MPa[/tex]
Now,
Direction of applied force: [0 1 0]
Plane: [1 0 1]
Direction: [tex][\bar{1}\ 1\ 1][/tex]
[tex]cos\theta = \frac{(0\times 1) + (1\times 0) + (0\times 1)}{1\times \sqrt{1^{2} + 1^{2}}} = 0[/tex]
[tex]cos\lambda = \frac{(0\times - 1) + (1\times 1) + (0\times 1)}{1\times \sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}[/tex]
[tex]\sigma_{R} = \sigma cos\theta cos\lambda = 2.75\times 0\times \frac{1}{\sqrt{3}}[/tex]
[tex]\sigma_{R} = 0 MPa[/tex]
Now,
It is clear that for the first case, the resolved shear stress is higher.
Thus the most favorable slip option is for [1 1 0] plane.
