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 [1¯11] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.8 MPa.

Answer 1

Stress is 7.716 MPa

Solution:

As per the question:

Magnitude of the resolved shear stress,
`\sigma_(R) = 2.8\ MPa`

No. of atoms in Body Centered Cubic Crystal, n = 2

Now,

Shear stress is given by:

`\sigma = (\sigma_(R))/(cos\theta cos\lambda)`

where

`\theta` = angle between slip plane and the tensile axis

`\lambda` = angle between tensile axis and the slip direction, i.e., [1 2 1] and
`[\bar{1}\ 1\ 1]`

Direction of applied force: [1 2 1]

Plane: [1 0 1]

Direction:
`[\bar{1}\ 1\ 1]`

Also, we know that:

`cos\lambda = \frac{(1* - 1) + (2* 1) + (1* 1)}{\sqrt{1^(2) + 2^(2) + 1^(2)}* \sqrt{(- 1)^(2) + 1^(2) + 1^(2)}} = (2)/(3√(2))`

Now,

`cos\theta = \frac{(1* 1) + (2* 0) + (1* 1)}{\sqrt{1^(2) + 2^(2) + 1^(2)}* \sqrt{(1)^(2) + 0^(2) + 1^(2)}} = (2)/(2√(3))`

Now,

Stress,
`\sigma = (2.8)/((2)/(2√(3))* (2)/(3√(2)))`

`\sigma = 7.716\ MPa`