Answer:
Stress is 7.716 MPa
Solution:
As per the question:
Magnitude of the resolved shear stress, [tex]\sigma_{R} = 2.8\ MPa[/tex]
No. of atoms in Body Centered Cubic Crystal, n = 2
Now,
Shear stress is given by:
[tex]\sigma = \frac{\sigma_{R}}{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, i.e., [1 2 1] and [tex][\bar{1}\ 1\ 1][/tex]
Direction of applied force: [1 2 1]
Plane: [1 0 1]
Direction: [tex][\bar{1}\ 1\ 1][/tex]
Also, we know that:
[tex]cos\lambda = \frac{(1\times - 1) + (2\times 1) + (1\times 1)}{\sqrt{1^{2} + 2^{2} + 1^{2}}\times \sqrt{(- 1)^{2} + 1^{2} + 1^{2}}} = \frac{2}{3\sqrt{2}}[/tex]
Now,
[tex]cos\theta = \frac{(1\times 1) + (2\times 0) + (1\times 1)}{\sqrt{1^{2} + 2^{2} + 1^{2}}\times \sqrt{(1)^{2} + 0^{2} + 1^{2}}} = \frac{2}{2\sqrt{3}}[/tex]
Now,
Stress, [tex]\sigma = \frac{2.8}{\frac{2}{2\sqrt{3}}\times \frac{2}{3\sqrt{2}}}[/tex]
[tex]\sigma = 7.716\ MPa[/tex]
