Question

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.

Answer 1

Step-by-step explanation:

From the given information:

The equation for applied stress can be expressed as:

`\sigma_(app) = (\tau_(CRSS))/(cos \phi \ cos \lambda)`

where;

`\phi` = angle between the applied stress [100] and [111]

To determine the
`\phi` and
`\lambda` for the system

Using the equation:

`\phi= cos^(-1)\Big [(l_1l_2+m_1m_2+n_1n_2)/(√((l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)))\Big]`

for [100]

`l_1 = 1, m_1 = 0, n_1 = 0`

for [111]

`l_1 = 1 , m_1 = 1, n_1 = 1`

Thus;

`\phi= cos^(-1)\Big [(1*1+0*1+0*1)/(√((1^2+0^2+0^2)(1^2+1^2+1^2)))\Big]`

`\phi= cos^(-1)\Big [(1)/(√((3)))\Big]`

`\phi= 54.74^0`

To determine
`\lambda` for
`[1 \overline 1 0]`

where;

for [100]

`l_1 = 1, m_1 = 0, n_1 = 0`

for
`[1 \overline 1 0]`

`l_1 = 1 , m_1 = -1, n_1 = 0`

Thus;

`\lambda= cos^(-1)\Big [(1*1+0*1+0*0)/(√((1^2+0^2+0^2)(1^2+(-1)^2+0^2)))\Big]`

`\phi= cos^(-1)\Big [(1)/(√((2)))\Big]`

`\phi= 45^0`

Thus, the magnitude of the applied stress can be computed as:

`\sigma_(app) = (\tau_(CRSS))/(cos \phi \ cos \lambda )`

`\sigma_(app) = (2.00)/(cos (54.74) \ cos (45) )`

`\mathbf{\sigma_(app) =4.89 \ MPa}`